5 Congruence of Segments, Angles and Triangles

This section deals with David Hilbert’s axiomatization of neutral geometry, and followsquite closely Hilbert’s foundations of geometry.

5.1 Congruence of segments

Proposition 5.1 (Congruence is an equivalence relation). Congruence is an equiv-alence relation on the class of line segments.

Question. Which three properties do we need to check for a congruence relation?

Answer. For an arbitrary relation to be a congruence relation, we need to check

(a) reflexivity: Each segment is congruent to itself, written AB ∼= AB.

(b) symmetry: If AB ∼= A′B′, then A′B′ ∼= AB.

(c) transitivity: If AB ∼= CD, and CD ∼= EF , then AB ∼= EF .

Proof of reflexivity, for a bottle of wine from Hilbert personally. 23 We need to show thateach segment is congruent to itself. Take any given segment AB. We transfer the seg-ment to a ray starting from any point C. By Hilbert’s axiom III.1, there exists a pointD on that ray such that AB ∼= CD. Now use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

Hence, for a case with other notation, AB ∼= CD and AB ∼= CD imply AB ∼= AB.

Proof of symmetry. Assume that AB ∼= CD. Because of reflexivity CD ∼= CD. Oncemore, we use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

Hence CD ∼= CD and AB ∼= CD imply CD ∼= AB.

Proof of transitivity. Assume AB ∼= CD and CD ∼= EF . Because of symmetry EF ∼=CD. Now we use Hilbert’s axiom III.2:

"If A′B′ ∼= AB and A′′B′′ ∼= AB, then A′B′ ∼= A′′B′′."

For a case with other notation, this means that AB ∼= CD and EF ∼= CD implyAB ∼= EF , as to be shown.

23which one can take gracefully as a thank-you—for translating from page 15 of the millenium editionof ”Grundlagen der Geometrie”–etc.

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Proposition 5.2 (Existence and uniqueness of segment transfer). Given a seg-ment AB and given a ray r originating at point A′, there exists a unique point B′ onthe ray r such that AB ∼= A′B′.

Question. Which part of this statement is among Hilbert’s axioms? Which axiom isused?

Answer. The existence of the segment A′B′ is postulated in Hilbert’s axiom of congru-ence III.1.

Question. How does the uniqueness of segment transfer follow? Which axioms areneeded for that part?

Answer. The uniqueness of segment transfer follows from the uniqueness of angle trans-fer, stated in III.4, and the SAS Axiom III.5.

Figure 5.1: Uniqueness of segment transfer

Proof of uniqueness. Assume the segment AB can be transferred to the ray r from A′

in two ways, such that both AB ∼= A′B′ and AB ∼= A′B′′. We choose a point C ′ not onthe line A′B′. One obtains the congruences

A′B′ ∼= A′B′′ , A′C ′ ∼= A′C ′ , ∠B′A′C ′ ∼= ∠B′′A′C ′

We are using axiom (III.2), the fact that a segment is congruent to itself, and an angle iscongruent to itself by the last statement of axiom (III.4). By the axiom axiom III.5 forSAS, this implies ∠A′C ′B′ ∼= ∠A′C ′B′′. By the uniqueness of angle transfer, as stated

in axiom III.4, this implies that the rays−−→C ′B′ =

−−−→C ′B′′ are equal. Hence B′ = B′′ is the

unique intersection point of the two different rays r =−−→A′B′ and

−−→C ′B′. We have shown

that AB ∼= A′B′ and AB ∼= A′B′′ imply B′ = B′′. Thus segment transfer is unique.

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Proposition 5.3 (Subtraction of segments). Given are three points on a line suchthat A ∗B ∗ C, and two points B′ and C ′ on a ray emanating from A′. Suppose that

AB ∼= A′B′ , AC ∼= A′C ′

then BC ∼= B′C ′ and A′ ∗B′ ∗ C ′ follow.

Figure 5.2: Segment subtraction

Proof. On the ray originating from B′, opposite to−−→B′A′, we transfer segment BC. We

get the seventh point S such that BC ∼= B′S, and A′ ∗B′ ∗ S. By Hilbert’s axiom III.3on segment addition, AB ∼= A′B′ and BC ∼= B′S imply now AC ∼= A′S.

On the other hand, it is assumed that AC ∼= A′C ′. At first, it seems that the locationof C ′ could be like C ′? or C ′′? in the drawing. But notice, as explained in problem 0.2:the transfer of segment AC is unique. Hence we get C ′ = S. Finally BC ∼= B′S andA′ ∗B′ ∗ S imply BC ∼= B′C ′ and A′ ∗B′ ∗ C ′, as to be shown.

Definition 5.1 (Segment comparison). For any two given segments AB and CDwe say that AB is less than CD, iff there exists a point E between C and D such thatAB ∼= CE and C ∗E ∗D. In this case, also say that CD is greater than AB. We writeCD > AB or AB < CD. equivalently.

Proposition 5.4 (Segment comparison holds for congruence classes). AssumingAB ∼= A′B′ and CD ∼= C ′D′, we get: AB < CD if and only if A′B′ < C ′D′.

Figure 5.3: Segment comparison for congruence classes

Proof. Transfer segment AB onto ray−−→CD, and get AB ∼= CE. The assumption AB <

CD implies that C ∗ E ∗ D. Too, we transfer segment A′B′ onto ray−−→C ′D′, and get

A′B′ ∼= C ′E ′.

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We now use segment subtraction for points C,E,D and C ′, E ′, D′. Since CD ∼= C ′D′

by assumption, and CE ∼= AB ∼= A′B′ ∼= C ′E ′—by assumption and construction andtransitivity—, and C ∗E ∗D; segment subtraction yields ED ∼= E ′D′ and C ′ ∗E ′ ∗D′.Thus E ′ lies between C ′ and D′, from which we conclude A′B′ < E ′D′.

Proposition 5.5 (Transitivity of segment comparison). If AB < CD and CD < EF ,then AB < EF .

Proof. I can assume that all three segments lie on the same ray−→AB, and A = C = E.

(Some transferring of segments will produce segments congruent with the given oneswhich satisfy these requirements.) Now, having done this, we get AB < AD andAD < AF . By definition this means A ∗B ∗D and A ∗D ∗ F .

Question. Express these order relations in words.

Answer. Point B lies between A and D, and point D lies between A and F .

By Theorem 5 from Hilbert’s foundations, any four points on a line can be notatedin a way that all four alphabetic order relations hold. (I shall that statement the ”Fourpoint Theorem”). Now the four points A,B,D, F satisfy the order relations A ∗ B ∗Dand A ∗D ∗ F , which implies they are already notated in alphabetic order. Hence thetwo other order relations follow: A ∗ B ∗ F and B ∗D ∗ F . But A ∗ B ∗ F means bydefinition that AB < AF , as to be shown.

Figure 5.4: Transitivity of segment comparison

Corollary 4. If AB < AD and AD < AF , then BD < BF .

Proposition 5.6 (Any two segments are comparable). For any two segments ABand CD, one and only one of the three cases (i)(ii)(iii) occurs:

Either (i) AB < CD or (ii) AB ∼= CD or (iii) CD < AB.

Proof. Transfer segment AB onto the ray−−→CD. We get AB ∼= CB′. Now the two

segments CB′ and CD start at the same vertex and lie on the same ray.By Theorem 4 from Hilbert’s foundations, for any given three points on a line, exactly

one lies between the other two. Hence the three points C,B′, D can satisfy

either (i) C ∗ B′ ∗ D. or (ii) B′ = D. or (iii) C ∗ D ∗ B′. These cases

correspond to the three cases as claimed. Indeed, in case (i), C ∗ B′ ∗ D implies bydefinition AB < CD. Indeed, in case (ii), B′ = D implies AB ∼= CB′ = CD by

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Figure 5.5: All segments are comparable

construction. An explanation may be needed in case (iii): One transfers segment CD

back onto the ray−→AB and gets CD ∼= AD0. Since CB′ ∼= AB, we can now use segment

subtraction, and get DB′ ∼= D0B and A ∗ D0 ∗ B. Because of the last order relation,

transferring segment CD back to ray−→AB confirms CD < AB.

Figure 5.6: To extend a line—.

Lemma 5.1 (”simple fact”). Given a segment AB and a ray−−→CD, there exists a point

R on this ray such that the segment DR is longer than the segment AB.

Proof. By the axiom of order (II.2), there exists a point P such that C ∗D ∗P . By theaxiom of congruence (III.1), there exists a point Q on the line CD, lying on the sameside of D as point P , such that AB ∼= DQ. By the axiom of order (II.2), there exists apoint R such that P ∗Q ∗R.

In the drawing on page 136, the given ray−−→CD is extended to obtain a point P outside

the segment CD. Next we transfer the given segment AB onto the extension ray−−→DP .

Finally, the ray−→PQ can be extended even more to get a point R outside the segment

PQ.The construction has produced points C,D,Q,R on a line satisfying the first two

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order relations below. We invoke the Four point Theorem, Proposition 3.3.

C∗D∗Q and

D∗Q∗R imply

C∗D∗ R

The third order relation confirms that point R lies on the ray−−→CD. The order relation

D ∗Q∗R means that the segment DR is longer than the segment DQ. By construction,the segments AB ∼= DQ are congruent. We see that

DR > DQ ∼= AB

By Proposition 5.4, segment comparison holds for congruence classes. We concludeDR > AB, as required.

Remark. What does Euclid’s Second Postulate ”To extend a line” really mean?A bid of thought shows that this postulate means more than just Hilbert’s axiom

of order (II.2). Remember Achilles and the tortoise! Indeed, we often need to extenda given line beyond a given segment. This process involves a combination of the axiomof order (II.2) with the axiom of congruence (III.1) for the segment transfer.

Another situation occurs when one knows, for any other reason, some point E beyondwhich one has to extend the line. Even in that situation, we need to invoke the Four-point Theorem (Hilbert’s Proposition 5) for the possibility of further extension beyondE. So we end up with different possible interpretations of Euclid’s postulate!

Definition 5.2 (Sum of segments). The sum of any two segments AB and EF isdefined to be the segment AC were C is the point such that A ∗B ∗ C and EF ∼= BC.

In other words, the sum segment AB +EF is obtained by extending the ray−→AB to

the point C by a segment BC congruent to the second summand EF .

Proposition 5.7 (Congruence classes of segments are an ordered Abeliangroup). The sum of segments is defined on equivalence classes of congruent segments.These equivalence classes have the following properties:

Commutativity a+ b = b+ a.

Associativity (a+ b) + c = a+ (b+ c)

Comparison Any two segments a and c satisfy either a < c, or a = c or a > c.

Difference a < c if and only if there exists b such that a+ b = c.

Comparison of Sums If a < b and c < d, then a+ c < b+ d.

Hence theses equivalence classes are an ordered Abelian group.

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Proof. By Hilbert’s axiom of congruence III.3, the sum is defined on equivalence classesof congruent segments. Now we check the items stated:

Commutativity: Let segment AB represent the equivalence class a. By axiom III.1,we can choose point C such that A ∗ B ∗ C, and segment BC represents theequivalence class b. Since BC = CB, and BA = AB, these two segments representthe classes b and a. By the definition of segment addition CA = CB + BA, andthe segment represents b+ a. Hence AC = CA implies that a+ b = b+ a.

Associativity: We construct (a + b) + c. To this end, choose a segment AB of con-gruence class a, in short AB ∈ a. Again, choose point C such that A ∗B ∗C, andsegment BC ∈ b. Furthermore, choose point D such that A ∗C ∗D, and segmentCD ∈ c. We now know that

AD ∈ (a+ b) + c

On the other hand, we begin by constructing at first b + c. By the four-pointtheorem, A ∗ B ∗ C and A ∗ C ∗ D imply B ∗ C ∗ D. Hence BD ∈ b + c. Theconstruction of a+ (b+ c) is finished by finding point H such that A ∗B ∗H, andsegment BH ∈ b+ c. We now know that

AH ∈ a+ (b+ c)

The uniqueness of segment transfer implies H = D. Finally, we see that AD =AH, and hence (a+ b) + c = a+ (b+ c).

Comparison: Let any segments a and c be given. These equivalence classes can be

represented by segments AB ∈ a and AC ∈ c on the same ray−→AB =

−→AC. The

three point theorem implies that either B = C, or A ∗B ∗C, or A ∗C ∗B. Henceeither a = c, or a < c, or c < a. In the last case a > c as shown above.

Difference: Assume a < c. These equivalence classes can be represented by segmentsAB ∈ a and AC ∈ c such that A ∗ B ∗ C. Now BC represents the class b suchthat a+ b = c. The converse is as obvious.

Comparison of Sums The proof is left to the reader.

Proposition 5.8 (Comparison of supplements). Let a segment PQ and two pointsA,B in it be given. Assume that PA < PB. Then BQ < AQ.

Proof. The reader has to try himself.

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5.2 Some elementary triangle congruences

Definition 5.3 (Triangle, Euler’s notation). For a triangle �ABC, it is assumedthat the three vertices do not lie on a line. I follow Euler’s conventional notation forvertices, sides and angles: in triangle �ABC, let A, B and C be the vertices, let thesegments a = BC, b = AC, and c = AB be the sides and the angles α := ∠BAC,β := ∠ABC, and γ := ∠ACB be the angles.

For a segment AB, it is assumed that the two endpoints A and B are different. Fora triangle �ABC, it is assumed that the three vertices do not lie on a line.

Proposition 5.9 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert’s theo-rem 11] An isosceles triangle has congruent base angles.

Question. Formulate the theorem with specific quantities from a triangle �ABC. Pro-vide a drawing.

Answer. If a ∼= b, then α ∼= β.

Figure 5.7: An isosceles triangle

Proof. This is an easy application of SAS-congruence. Assume that the sides AC ∼= BCare congruent in �ABC. We need to show that the base angles α = ∠BAC andβ = ∠ABC are congruent. Define a second triangle �A′B′C ′ by setting

A′ := B , B′ := A , C ′ := C

(It does not matter that the second triangle is just ”on top” of the first one.) To applySAS congruence, we match corresponding pieces:

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(1) ∠ACB = ∠BCA = ∠A′C ′B′ because the order of the sides of an angle is arbitrary.By axiom III.4, last part, an angle is congruent to itself. Hence ∠ACB ∼= ∠A′C ′B′.

(2) AC ∼= A′C ′.

Question. Explain why this holds.

Answer. AC ∼= BC because we have assumed the triangle to be isosceles, andBC = A′C ′ by construction. Hence AC ∼= A′C ′.

(3) Similarly, we show that (3): BC ∼= B′C ′: Indeed, BC ∼= AC because we haveassumed the triangle to be isosceles, and congruence is symmetric, and AC = B′C ′

by construction. Hence BC ∼= B′C ′.

Finally, we use axiom III.5. Items (1)(2)(3) imply ∠BAC ∼= ∠B′A′C ′ = ∠ABC. Butthis is just the claimed congruence of base angles.

The next Proposition is Theorem 12 of Hilbert’s Foundations of Geometry: Hereonly the weak form of the SAS-axiom (Hilbert’s Axiom III.5) is assumed.

Question. Why is Hilbert’s Axiom III.5 weaker than the SAS congruence theorem?

Answer. In Hilbert’s Axiom, only congruence of a further pair of angles is postulated.In the SAS congruence theorem, all pieces of the two triangles are stated to be pairwisecongruent.

Figure 5.8: SAS congruence

Proposition 5.10 (Hilbert’s SAS-axiom implies the full SAS Congruence The-orem). [Theorem 12 in Hilbert] Given are two triangles. We assume that two sides andthe angle between these sides are congruent to the corresponding pieces of the second tri-angle. Then the two triangles are congruent, which means that all corresponding piecesare pairwise congruent.

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Proof. Let �ABC and �A′B′C ′ be the two triangles. We assume that the angles at Aand A′ as well as two pairs of adjacent sides are matched:

c = AB ∼= A′B′ = c′ , b = AC ∼= A′C ′ = b′ , α = ∠CAB ∼= ∠C ′A′B′ = α′

We need to show that

β = ∠ABC∼=∠A′B′C ′ = β′(a)

γ = ∠BCA∼=∠B′C ′A′ = γ′(b)

a = BC∼=B′C ′ = a′(c)

Note that by Hilbert’s weaker SAS-axiom, we can concluded only part (a). Part (b)follows immediately by applying Hilbert’s axiom to triangles �ACB and �A′C ′B′. We

need still to show part (c). Transferring segment BC onto ray−−→B′C ′, we get a seventh

point D′ on that ray such that

(1) BC ∼= B′D′

Applying Hilbert’s (weak) SAS axiom to the two triangles �ABC and �A′B′D′ yields∠BAC ∼= ∠B′A′D′. On the other hand, by assumption ∠BAC ∼= ∠B′A′C ′. By unique-ness of angle transfer (see Hilbert’s axiom III.4), there exists one and only one ray, that

forms with the ray−−→A′B′ the given ∠BAC, and lies on the same side of the line A′B′ as

point C ′. Hence rays−−→A′D′ =

−−→A′C ′ are equal, in other words D′ lies on the ray

−−→A′C ′.

By construction of �A′B′D′, point D′ lies on the ray−−→B′C ′, too. Since the intersection

point of the two lines A′C ′ and B′C ′ is unique, one concludes

(2) C ′ = D′

From (1) and (2) we get BC ∼= B′C ′, as to be shown.

Proposition 5.11 (Extended ASA-Congruence Theorem). Given is a triangleand a segment congruent to one of its sides. The two angles at the vertices of this sideare transferred to the endpoints of the segment, and reproduced in the same half plane.Then the newly constructed rays intersect, and one gets a second congruent triangle.

Proof. Let the triangle be �ABC and the segment A′B′ ∼= AB. The angle β = ∠ABCis transferred at B′ along ray

−−→B′A′, and the angle α = ∠BAC is transferred at A′ along

ray−−→A′B′, both are reproduced on the same side of line A′B′. One gets two new rays r′A

and r′B such that

α = ∠(−→AB,

−→AC) ∼= ∠(

−−→A′B′, r′A) , β = ∠(

−→BA,

−−→BC) ∼= ∠(

−−→B′A′, r′B)

It is claimed that the two new rays r′A and r′B do intersect at some point C ′′. Further-more, it is claimed that the two triangles �ABC and �A′B′C ′′ are congruent.

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Figure 5.9: Extended ASA congruence

On the newly produced ray r′A, we transfer segment AC and get a point C ′′ suchthat AC ∼= A′C ′′. Now SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10above) implies

(1) �ABC ∼= �A′B′C ′′

The three pairs of matching pieces used to prove the congruence are stressed in matching

colors. Congruence (1) implies β = ∠ABC ∼= ∠A′B′C ′′ = ∠(−−→B′A′,

−−−→B′C ′′). On the other

Figure 5.10: How to simply get ASA congruence

hand, ∠ABC ∼= ∠(−−→B′A′, r′B) was assumed, too. Hence by uniqueness of angle transfer,

we get two equal rays:−−−→B′C ′′ = r′B. Thus the point C ′′ lies on the newly produced ray

r′B, too. Thus the two rays r′A and r′B intersect in point C ′′, and (1) holds, as to beshown.

Proposition 5.12 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles witha pair of congruent sides, and pairwise congruent adjacent angles are congruent.

Problem 5.1. State the ASA congruence, with the notation from the figure above. Provethe theorem.

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Answer (Independent proof of ASA congruence). Given are the triangles �ABC and�A′B′C ′, with congruent sides AB ∼= A′B′ and two pairs of congruent adjacent angles

at A,A′ and B,B′. The segment AC is transferred onto the ray−−→A′C ′. On this ray, one

gets a point X such that AC ∼= A′X. Now SAS congruence is applied to the triangles�ABC and �A′B′X. The three pairs of matching pieces used to prove the congruenceare stressed in matching colors. By axiom (III.5), one concludes

(5.1) ∠ABC ∼= ∠A′B′X

and from the SAS congruence theorem one gets even

�ABC ∼= �A′B′X(5.2)

We now have obtained both ∠ABC ∼= ∠A′B′X, and ∠ABC ∼= ∠A′B′C ′ as assumed,too. Hence by uniqueness of angle transfer, we know the rays to be equal:

−−→B′X =

−−→B′C ′

Thus the point X lies on both this ray, and the ray−−→A′X =

−−→A′C ′, too. These two rays

have a unique intersection point, since they do not lie on the same line. Hence X = C ′,and (5.2) is just the required triangle congruence.

Problem 5.2. In which point does the extended ASA congruence theorem extend theusual ASA congruence theorem?

Answer. The extended ASA theorem differs from the usual ASA-congruence theorem,because existence of a second triangle is not assumed—besides the first triangle, only asecond segment is given. It is proved that the two newly produced rays do intersect.

Proposition 5.13 (Preliminary Converse Isosceles Triangle Proposition). If the twobase angles of a triangle are congruent to each other, the triangle is isosceles.

Question. Formulate the theorem with specific quantities from a triangle �ABC. Pro-vide a drawing.

Answer. If α ∼= β and β ∼= α, then a ∼= b.

Remark. We can avoid that awkward assumption, ”If α ∼= β and β ∼= α”, later inProposition 5.32. Hilbert proves the converse isosceles triangle theorem as his Theorem24, after having the exterior angle theorem at his disposal.

Proof. We use ASA-congruence to prove the Proposition. Assume that the angles α =∠CAB and β = ∠ABC are congruent in �ABC. We need to show that the two sidesa = BC and b = AC are congruent. Define a second triangle �A′B′C ′ by setting

A′ := B , B′ := A , C ′ := C

(It does not matter that the second triangle is just ”on top” of the first one.) To applyASA congruence, we match corresponding pieces:

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Figure 5.11: An isosceles triangle, two ways to look at it

(1) AB = B′A′ = A′B′. Hence AB ∼= A′B′, because the order of the endpoints of asegment is arbitrary, and a segment is congruent to itself.

(2) α ∼= α′.

Question. Explain why this holds.

Answer. α ∼= β by assumption, and β = ∠ABC = ∠B′A′C ′ = α′ by construction.Hence α ∼= α′.

(3) Similarly, one shows that β ∼= β′: β ∼= α by assumption, and α = ∠BAC =∠A′B′C ′ = β′ by construction. Hence β ∼= β′.

Via ASA congruence, items (1)(2)(3) imply that �ABC ∼= �A′B′C ′, and hence espe-cially AC ∼= A′C ′ = BC as to be shown.

5.3 Congruence of angles

Definition 5.4 (Supplementary Angles). Two angles are called supplementary an-gles, iff they have a common vertex, both have one side on a common ray, and the twoother sides are the opposite rays on a line.

Definition 5.5 (Vertical Angles). Two angles are called vertical angles, iff they havea common vertex, and their sides are two pairs of opposite rays on two lines.

Proposition 5.14 (Congruence of Supplementary Angles). [Theorem 14 of Hilbert]If an angle ∠ABC is congruent to another angle ∠A′B′C ′, then its supplementary angle∠CBD is congruent to the supplementary angle ∠C ′B′D′ of the second angle.

Proof. The three steps of the proof each identify a new pair of congruent triangles.

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Step 1: One can choose the points A′, C ′ and D′ on the given rays from B such that

AB ∼= A′B′ , CB ∼= C ′B′ , DB ∼= D′B′

Because of the assumption ∠ABC ∼= ∠A′B′C ′, SAS congruence (Theorem 12 of Hilbert,see Proposition 5.10 above) now implies that �ABC ∼= �A′B′C ′. In the drawing, thethree pairs of matching pieces used to prove the congruence are stressed in matchingcolors. Congruence of the two triangles implies

Figure 5.12: Congruence of supplementary angles, the first pair of congruent triangles

(1) AC ∼= A′C ′ and ∠BAC ∼= ∠B′A′C ′

Step 2: By axiom III.3, adding congruent segments yields congruent segments. Hencethe segments AD and A′D′ are congruent. Now SAS congruence (Theorem 12 of Hilbert,see Proposition 5.10 above) implies that the (greater) triangles, too, are congruent inthe two figures below. In the drawing, the three pairs of matching pieces used to provethe congruence are stressed in matching colors. The congruence �CAD ∼= �C ′A′D′

Figure 5.13: Congruence of supplementary angles, the second pair of congruent triangles

implies

(2) CD ∼= C ′D′ and ∠ADC ∼= ∠A′D′C ′

Step 3: At last we consider the two triangles �BCD and �B′C ′D′ on the right side.In the drawing, the three pairs of matching pieces used to prove the congruence arestressed in matching colors. Again by using SAS congruence (Theorem 12 of Hilbert,see Proposition 5.10 above), we see the two triangles are congruent. Finally ∠CBD ∼=∠C ′B′D′, as to be shown.

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Figure 5.14: Congruence of supplementary angles, the third pair of congruent triangles

Figure 5.15: Supplementary angles yield points on a line

Corollary 5. Adjacent angles congruent to supplementary angles are supplementary,too.

Proof. Given are supplementary angles ∠ABC and ∠DBC, a further congruent angle∠ABC ∼= ∠A′B′C ′, and a point D′ with A′ and D′ lying on different sides of line B′C ′.

We shown that the angles ∠A′B′C ′ and ∠D′B′C ′ are supplementary if and only if∠CBD ∼= ∠C ′B′D′. Above, we have already shown one direction: If the angles ∠A′B′C ′and ∠D′B′C ′ are supplementary, then ∠CBD ∼= ∠C ′B′D′.

Now we show the converse. Assume that ∠CBD ∼= ∠C ′B′D′. We have to checkwhether point B′ lies between A′ and D′. Choose any point D′′ on the ray opposite to−−→B′A′. Congruence of supplementary angles (Hilbert’s Theorem 14, see Proposition 5.14above) implies ∠CBD ∼= ∠C ′B′D′′. On the other hand, ∠CBD ∼= ∠C ′B′D′ is assumed.

Angle ∠CBD is transferred uniquely along ray−−→B′C into the half plane opposite to A′.

Indeed, by axiom III.4, angle transfer produces a unique new ray. Hence−−−→B′D′′ =

−−→B′D′.

and the four points A′, B′, D′, D′′ lie on one line, with point B′ between A′ and D′.Hence angles ∠A′B′C ′ and ∠D′B′C ′ are supplementary, too.

Proposition 5.15 (Congruence of Vertical Angles). [Euclid I.15] Vertical anglesare congruent.

Proof. This is an easy consequence of Hilbert’s Theorem 14 about supplementary angles(see Proposition 5.14 above). Take any two vertical angles ∠ABC and ∠A′BC ′. Weassume that vertex B lies between points A and A′ on one line, as well as betweenthe two points C and C ′ on a second line. As shown in the figures, angle ∠ABC ′ hasangle (i) ∠ABC as supplementary angle. Secondly, angle ∠ABC ′ has (ii) ∠A′BC ′ assupplementary angle, too.

An angle is congruent to itself, as stated in Axiom III.4. Hence especially ∠ABC ′ ∼=∠ABC ′. By Theorem 14 of Hilbert (see Proposition 5.14 above), angles supplementary

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Figure 5.16: Two pairs of supplementary angles yield vertical angles

to congruent angles are congruent, too. Hence we conclude congruence of the two verticalangles: ∠ABC ∼= ∠A′BC ′, as to be shown.

Proposition 5.16 (The main case of angle-addition). Given is an angle ∠ABCand a ray

−−→BG in its interior, as well as a second angle ∠A′B′C ′ and a ray

−−→B′G′ in its

interior. Furthermore, assume that ∠CBG ∼= ∠C ′B′G′ and ∠ABG ∼= ∠A′B′G′. Thenthe two angle sums are congruent, too: ∠ABC ∼= ∠A′B′C ′.

Figure 5.17: Angle addition

Proof. By the Crossbar Theorem, a segment going from one side of an angle to theother, and a ray in the interior of that angle always intersect. Hence there exists a point

H such that−−→BG =

−−→BH and A � H � C. For simplicity, we choose G = H from the

beginning. Too, we may assume that A′, C ′ and G′ are chosen such that BA ∼= B′A′,BC ∼= B′C ′ and BG ∼= B′G′. The proof uses three pairs of congruent triangles, in step(1)(2)(3), respectively.

Step (1): The SAS-congruence axiom implies �BGC ∼= �B′G′C ′ because of∠GBC ∼= ∠G′B′C ′ , GB ∼= G′B′ and BC ∼= B′C ′ which hold by assumption and the

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remarks above. In the drawing, the three pairs of matching pieces used to prove thecongruence are stressed in matching colors. From the congruence of these two triangles,

Figure 5.18: the first pair of congruent triangles

we conclude that

∠BCG ∼= ∠B′C ′G′(1)

∠BGC ∼= ∠B′G′C ′(1b)

Step (2): As a second step, the SAS-congruence axiom implies �BGA ∼= �B′G′A′because of ∠GBA ∼= ∠G′B′A′ , GB ∼= G′B′ and BA ∼= B′A′ which hold by assumptionand the remarks above. Again, in the drawing, the three pairs of matching pieces usedto prove the congruence are stressed in matching colors. By construction, the two angles

Figure 5.19: the second pair of congruent triangles

∠BGA and ∠BGC are supplementary angles. The congruence ∠BGA ∼= ∠B′G′A′ fol-lows from step (2), and ∠BGC ∼= ∠B′G′C ′ as stated by (1b). As derived in our Corollaryto Hilbert’s Theorem 14 (see Proposition 5.14 above), adjacent angles congruent to sup-plementary angles are supplementary, too. Hence the two angles ∠B′G′A′ and ∠B′G′C ′

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are supplementary. Hence the three points A′, G′ and C ′ lie on a straight line. Bysegment addition (Hilbert’s axiom III.3), AG ∼= A′G′ and GC ∼= G′C ′ imply

(2) AC ∼= A′C ′

Step (3): To set up a third pair of congruent triangles, we use (1) (2) and the assumption

(3) BC ∼= B′C ′

The SAS axiom shows that �ABC ∼= �A′B′C ′ because of (1)(2)(3). I have stressedthese three pairs of matching pieces in matching colors. Finally, ∠ABC ∼= ∠A′B′C ′

Figure 5.20: the third pair of congruent triangles

follows from �ABC ∼= �A′B′C ′, as to be shown.

Proposition 5.17 (The main case of angle subtraction). Given is an angle ∠ABCand a ray

−−→BG in its interior, as well as a second angle ∠A′B′C ′ and a ray

−−→B′G′ in its

interior. If ∠CBG ∼= ∠C ′B′G′ and ∠ABC ∼= ∠A′B′C ′ then ∠ABG ∼= ∠A′B′G′.

Proposition 5.18 (Angle-Addition and Subtraction). [Theorem 15 in Hilbert]Given are three vertices h, k, l with the common vertex O lying in a plane a, and threevertices h′, k′, l′ with the common vertex O′ lying in a plane a′. We assume either thecase of angle subtraction or angle addition.

angle subtraction: in this case the rays h and k lie in the same half-plane of l, andthe rays h′ and k′ lie in the same half-plane of l′;

angle addition: in this case the rays h and k lie in the different half-planes of l, andthe rays h′ and k′ lie in different half-planes of l′.

Given is an angle ∠ABC and a ray−−→BG in its interior, as well as a second angle ∠A′B′C ′

and a ray−−→B′G′ in its interior. If

∠(h, l) ∼= ∠(h′, l′) and ∠(l, k) ∼= ∠(l′, k′)

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then either both h and k are opposite rays, and h′ and k′ are opposite rays, too; or

∠(h, k) ∼= ∠(h′, k′)

Proof. We explain only the case of angle addition. In the special case that h and kare opposite rays, the angles ∠(h, l) and ∠(l, k) are adjacent supplements. Hence byCorollary 5, the congruent angles ∠(h′, l′) and ∠(l′, k′) are supplementary, too. In otherwords we get opposite rays h′ and k′. We have to distinguish the two remaining maincases:

(i) the two rays k and l lie in one half-plane of ray h;

(ii) the two rays k and l lie in different half-planes of ray h.

In the first case (i), the two rays k′ and l′ lie in one half-plane of ray h′. The resultfollows by angle addition as in Proposition 5.16.

Here is a detailed argument: We transfer angle ∠(h, k) onto the ray h′ in the half-plane determined by l′ and get a ray k′+ such that ∠(h, k) ∼= ∠(h′, k′+). The Proposi-tion 5.17 about the main case of angle subtraction implies ∠(l.k) ∼= ∠(l′, k′+). Becauseof the assumption ∠(l.k) ∼= ∠(l′, k′), uniqueness of angle transfer implies k′+ = k′ Hencethe two rays k′ and l′ lie in one half-plane of ray h′.

In the second case (ii), we use the opposition rays l− of l and l′− from l′. NowProposition 5.14 about the congruence of supplementary angles implies

∠(h, l−) ∼= ∠(h′, l′−) and ∠(l−, k) ∼= ∠(l′−, k′)

Since we are back to case (i), we can conclude the claim ∠(h, k) ∼= ∠(h′, k′).

Definition 5.6 (The sum of two angles). Let an angle ∠ABC and a ray−−→BG in its

interior. Angle ∠ABC is called the sum of the angles ∠ABG and ∠GBC. One writes∠ABC = ∠ABG+ ∠GBC.

Figure 5.21: Three cases for adding angles: they (i) are supplementary (ii) can be added(iii) or cannot be added

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Lemma 5.2. Given are two angles ∠ABG and ∠GBC with same vertex B lying on

different sides of a common ray−−→BG. Exactly one of three possibilities occur about the

angles ∠ABG and ∠GBC:

(i) They are supplementary. The three points A,B and C lie on a line.

(ii) They can be added. Their sum is ∠ABG+∠GBC = ∠ABC. Points A and Glie on the same side of line BC. Points C and G lie on the same side of line AB.

(iii) They cannot be added. Points A and G lie on different sides of line BC.Points C and G lie on different sides of line AB. (The sum would be an over-obtuse angle.)

Figure 5.22: If A and G lie on different sides of BC, then C and G lie on different sides ofAB.

Proof. Suppose that neither case (i) nor (ii) occurs. Under that assumption, either

(a) points A and G lie on different sides of line BC—or

(b) points C and G lie on different sides of line AB.

Suppose case (a) occurs. Segment AG intersects line BC, say in point H. We usePasch’s axiom for triangle �CHG and line AB. This line does not intersect side GH,but intersects side CH. Indeed, point B lies between C and H, since C and A, andhence C and H are on different sides of line BG.

Now Pasch’s axiom—for triangle �CHG and line AB—implies that this line inter-sects side CG, say at point Pasch. Thus points C and G lie on different sides of lineAB. We have shown case (iii) to occur.

Suppose case (b) occurs. The same argument—with A and C exchanged— showsthat case (iii) occurs once more.

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Remark. Because of these theorems, angle addition and subtraction are defined forcongruence classes of angles.

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5.4 SSS congruence

The ”Theorem about the symmetric kite” is needed as a preparation to get SSS con-gruence.

Proposition 5.19 (The Symmetric Kite). [Theorem 17 of Hilbert] Let Z1 and Z2 betwo points on different sides of line XY , and assume that XZ1 ∼= XZ2 and Y Z1 ∼= Y Z2.Under these assumptions

(i) the angles ∠XZ1Y ∼= ∠XZ2Y are congruent;

(ii) the angles ∠XY Z1 ∼= ∠XY Z2 are congruent.

Proof. The congruence of the base angles of isosceles�XZ1Z2 yields ∠XZ1Z2 ∼= ∠XZ2Z1.Similarly, one gets ∠Y Z1Z2 ∼= ∠Y Z2Z1. Now angle addition (or substraction) implies

(*) ∠XZ1Y ∼= ∠XZ2Y

Angle addition is needed in case ray−−−→Z1Z2 lies inside ∠XZ1Y , angle subtraction in case

ray−−−→Z1Z2 lies outside ∠XZ1Y .In the special case that either point X or Y lies on the line Z1Z2, one gets the same

conclusion even easier. Here are drawings for the three cases. One now applies SAS

Figure 5.23: The symmetric kite

congruence to the triangles �XZ1Y and �XZ2Y . Indeed, the angles at Z1 and Z2and the adjacent sides are pairwise congruent. Hence the assertion ∠XY Z1 ∼= ∠XY Z2follows.

Before getting the general SSS-congruence, I consider one further special case.

Lemma 5.3 (Lemma for SSS Congruence). Assume that the two triangles �ABCand �AB′C have a common side AC, and all three corresponding sides are congruent,and the two vertices B and B′ lie on the same side of line AC. Then the two trianglesare identical.

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Proof. We transfer the angle ∠BAC onto the ray−→AC, on the side of line AC opposite

to B and B′. On the newly produced ray, we transfer segment AB, starting at vertexA. Thus we get point B′′, and segment AB′′ ∼= AB. From SAS-congruence (Theorem12 of Hilbert, see Proposition 5.10 above), one concludes that BC ∼= B′′C. As stressedin the first drawing, one has constructed a symmetric kite with the four points

X := A, Y := C,Z1 := B,Z2 := B′′

Hence by Theorem 17 (see Proposition 5.19 above), ∠B′′AC ∼= ∠BAC

Figure 5.24: Which kite is symmetric?

But wait! Another choice of four points to get a kite is

X := A, Y := C,Z1 := B′, Z2 := B′′

—replacing B by B′. Because of the assumptions, and the construction of the first kite,AB′ ∼= AB ∼= AB′′ and B′C ∼= BC ∼= B′′C. We see that the second kite, stressed in thesecond drawing, satisfies the assumptions of Hilbert’s Theorem 17 (see Proposition 5.19above), too. Now we conclude from Hilbert’s Theorem 17 (see Proposition 5.19 above)

that ∠B′′AC ∼= ∠B′AC. Finally, the uniqueness of angle transfer implies that−→AB =−−→

AB′. Since AB ∼= AB′, and segment transfer was shown to produce a unique point, wehave confirmed that B = B′. Thus the two triangles �ABC and �AB′C are identical,as to be shown.

Remark. We do not need to assume that angle congruence is an equivalence relation.But therefore we have to consider two kites, by choosing two of the three triangles, asshown in two drawings. Then we can use that transferring an angle gives a unique ray.

Proposition 5.20 (The diagonals of the rhombus bisect each other perpen-dicularly). Given are four different points A,B,C,D such that the segments AB ∼=BC ∼= CD ∼= DA are congruent. Then the segments AC and BD bisect each otherperpendicularly at their common midpoint.

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Figure 5.25: The diagonals of the rhombus bisect each other.

Answer. Points A and C lie on different sides of line BD. Otherwise the lemma forSSS-congruence would imply A = C, contrary to the assumption that four differentpoints A,B,C,D are given. Similarly, we see that points B and D lie on different sidesof the other diagonal AC. Hence the two segments AC and BD intersect. Let point Mbe their intersection point.

We draw the horizontal diagonal AC. The upper- and lower triangles have two pairsof congruent bases angles (see second figure). Angle addition implies ∠ABD ∼= ∠CBD.By SAS congruence, the left- and right triangles �ABD ∼= �CBD are congruent (seethird figure).

Furthermore, the latter two triangles are both isosceles. Hence �ABD ∼= �DCBholds, too. Again by SAS congruence, we get four mutually congruent base angles withone side on the line BD (first figure in the second row). Similarly, we get four mutuallycongruent base angles with one side on the line AC.

By ASA congruence, we get �ABM ∼= �CDM . Hence the diagonals AC and BDbisect each other (see last figure in the second row).

Another ASA congruence yields �AMB ∼= �CMB. Hence the angles ∠AMB ∼=∠CMB are congruent supplements, and hence they are right angles. Hence the diagonalsAC and BD are perpendicular to each other.

From Hilbert’s Theorem 17 (see Proposition 5.19 above), and the Lemma 5.3 above,

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we see that SSS-congruence holds for any two triangles with a common side. Now wecan easily get the general case of

Proposition 5.21 (SSS Congruence). [Theorem 18 in Hilbert’s Foundations] Twotriangles with three pairs of congruent sides are congruent.

Figure 5.26: Which two triangles are congruent?

Proof. Assume the two triangles �ABC and �A′B′C ′ have corresponding sides which

are congruent. We transfer the angle ∠BAC onto the ray−−→A′C ′, at vertex A′, to the

same side of A′C ′ as B′. On the newly produced ray, we transfer segment AB, startingat vertex A′. Thus we get point B0, such that A′B0 ∼= AB. Because of SAS-congruence(Theorem 12 of Hilbert, see Proposition 5.10 above), we get

(*) �ABC ∼= �A′B0C ′

Hence especially, AB ∼= A′B0 ∼= A′B′ and BC ∼= B0C′ ∼= B′C ′. We can now apply

the Lemma‘5.3 to the two triangles �A′B′C ′ and �A′B0C ′. Hence B′ = B0, and theassertion follows from (*).

Corollary 6 (Transfer of a triangle). Any given triangle �ABC can be transferredinto both the left as well as right half-plane of any given ray r, to obtain a congruenttriangle with one vertex at the vertex of this ray, one side lying on the given ray, andthe third vertex in the prescribed half-plane.

5.5 The equivalence relation of angle congruence

Proposition 5.22 (Theorem 19 in Hilbert’s Foundations). If two angles ∠(h′, k′)and ∠(h′′, k′′) are congruent to a third angle ∠(h, k), then the two angles ∠(h′, k′) and∠(h′′, k′′) are congruent, too.

Proof for a bottle of wine from Hilbert to A. Rosenthal. 24 Let the vertices of the anglesbe O,O′, O′′. Choose points A,A′, A′′ on one side of the three angles, respectively,such that O′A′ ∼= OA and O′′A′′ ∼= OA. Similarly choose points B,B′, B′′ on the

24Hilbert gives credit for this proof to A. Rosenthal (Math. Ann. Band 71)

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three remaining sides, respectively, such that O′B′ ∼= OB and O′′B′′ ∼= OB. Now theassumption of SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10 above) aremet for both �A′O′B′ and �AOB, as well as �A′′O′′B′′ and �AOB. Hence

A′B′ ∼= AB , A′′B′′ ∼= AB

We use axiom III.2:

"two segments congruent to a third one are congruent to each other".

Hence �A′B′O′ and �A′′B′′O′′ have three pairs of congruent sides. By the SSS-congruence Theorem 18, we conclude that ∠(h′, k′) ∼= ∠(h′′, k′′), as to be shown.

Proposition 5.23 (Congruence is an equivalence relation). Congruence is anequivalence relation on the class of angles.

Question. Which three properties do we need to check for a congruence relation?

Answer. For an arbitrary relation to be a congruence relation, we need to check

(a) reflexivity: Each angle is congruent to itself, written α ∼= α.

(b) symmetry: If α ∼= β, then β ∼= α.

(c) transitivity: If α ∼= β, and β ∼= γ, then α ∼= γ.

Question. How can we claim reflexivity?

Answer. Reflexivity is given by the last part of axiom III.4

Proof of symmetry, for a bottle of wine from A. Rosenthal to Hilbert. 25 By Theorem 19(see Proposition 5.22 above), ∠(h′, k′) ∼= ∠(h, k) and ∠(h′′, k′′) ∼= ∠(h, k) imply ∠(h′, k′) ∼=∠(h′′, k′′). Hence, with just other notation, we see that β ∼= β and α ∼= β implyβ ∼= α.

Proof of transitivity. Assume α ∼= β and β ∼= γ. Because of symmetry γ ∼= β. Now weuse Theorem 19 (see Proposition 5.22 above). Hence, just with other notation, α ∼= βand γ ∼= β imply α ∼= γ.

Definition 5.7 (Angle comparison). Given are two angles ∠BAC and ∠B′A′C ′. We

say that ∠BAC is less than ∠B′A′C ′, iff there exists a ray−−→A′G in the interior of ∠B′A′C ′

such that ∠BAC ∼= ∠B′A′G. In this case, we also say that ∠B′A′C ′ is greater than∠BAC. We write ∠B′A′C ′ > ∠BAC and ∠BAC < ∠B′A′C ′, equivalently.

Proposition 5.24 (Angle comparison holds for congruence classes). Assumethat α ∼= α′ and β ∼= β′. α < β if and only if α′ < β′.

Proof. The reader should do it on her own.

25Hilbert gave credit and got back a good bottle

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Proposition 5.25 (Transitivity of angle comparison). If α < β and β < γ, thenα < γ.

Proof. After having done some transfer of angles, I can assume that all three angles

α, β, γ have the common side−→AB, and lie on the same side of AB. Choose any point E

on the second side of γ = ∠BAE.Because β < γ, the second side of β is in the interior of the largest angle γ. Hence,

by the Crossbar Theorem the segment BC intersects that second side of β, say at pointD, and β = ∠BAD as well as B ∗D ∗ E.

Because α < β, the second side of α is in the interior of β . Hence, by theCrossbar Theorem the segment BD intersects that second side of α, say at point E, andα = ∠BAE as well as B ∗ C ∗ D. Any four points on a line can be ordered in a way

Figure 5.27: Transitivity of comparison of angles

that all four alphabetic order relations hold. (see Theorem 5 in Hilbert, which I callthe ”four-point Theorem”). Now the four points B,C,D,E satisfy the order relationsB ∗ C ∗ D and B ∗ D ∗ E. Therefore they are already put in alphabetic order. HenceB ∗ C ∗ E . This shows by definition that α = ∠BAC < ∠BAE = γ.

Proposition 5.26 (All angles are comparable). For any two angles α and β, oneand only one of the three cases (i)(ii)(iii) occurs:

Either (i) α < β or (ii) α ∼= β or (iii) β < α.

Proof. Not more than one of the cases (i)(ii)(iii) can occur at once. Case (ii) excludeseither (i) or (iii) by definition of angle comparison. But case (i) and (iii) cannot holdboth at the same time, neither: By transitivity, α < β and β < α together would implyα < α. This is impossible, because an angle is congruent to itself: α ∼= α by axiom III.4.

Now we show that actually one of the cases (i)(ii)(iii) does occur. Transfer angle αonto one side of β. Thus we get the two angles α̃ = ∠BAC ∼= α and β = ∠BAD which

have the common side−→AB, and lie on the same side of AB. The second side

−→AC of

angle α̃ can

either (i) lie in the interior of angle β.

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or (ii) be identical to the second side of β.

or (iii) lie in the exterior of angle β.

These cases correspond to the three cases as claimed. An explanation may be needed in

Figure 5.28: All angles are comparable

case (iii): The ray−→AC lies in the exterior of angle β = ∠BAD. What does that mean?

Answer. Either points B and C lie on different sides of line AD, or points C and D lieon different sides of line AB.

The points C and D lie on the same side of AB by the arrangement of the angles. HenceB and C lie on different sides of AD. By the Crossbar theorem, segment BC intersects

ray−−→AD. In term to comparing angles, we get β = ∠BAD < ∠BAC = α̃ ∼= α.

Proposition 5.27 (Comparison of supplements). If α < β, then their supplementsS(α) and S(β) satisfy S(α) > S(β).

Figure 5.29: Comparison of supplements

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Proof. Transfer angle α onto one side of β, in the same half plane. Thus we get the two

angles α̃ ∼= α and β = ∠BAD which have the common side−→AB, and lie on the same

side of AB.Because of α < β, the second side of angle α̃ lies in the interior of angle β. By the

Crossbar theorem it intersects the segment BD, say at point Q. We have arranged that

α ∼= α̃ = ∠BAQ, and β = ∠BAD. Let F be any point on the ray opposite to−→AB.

Apply Pasch’s axiom (Hilbert’s axiom of order II.4) to �FBQ and line AD. That linedoes not intersect side BQ because of B ∗Q ∗D, but does intersect side FB because ofF ∗A∗B. Hence line AD intersects the third side FQ, say at point S. Indeed F ∗S ∗Q,

and S lies on the ray−−→AD because only that ray, and not its opposite ray lies in the

interior of ∠FAQ.The supplementary angle of β is S(β) = ∠FAD = ∠FAS. The supplementary angle

of α̃ is S(α̃) = ∠FAQ. From F ∗ S ∗Q we get

S(β) = ∠FAS = ∠FAD < ∠FAQ = S(α̃)

The final step uses Hilbert’s Theorem 14 (see Proposition 5.14 above): the supplementsof congruent angles are congruent, too. Hence α ∼= α̃ implies S(α) ∼= S(α̃). FromS(β) < S(α̃) and S(α) ∼= S(α̃), we conclude S(β) < S(α) as to be shown.

Definition 5.8 (acute, right and obtuse angles). A right angle is an angle congruentto its supplementary angle. An acute angle is an angle less than a right angle. An obtuseangle is an angle greater than a right angle.

Remark (Remark about supplements of acute and obtuse angles). If an angle α is acute,its supplement S(α) is obtuse. Furthermore, the angle is less than its supplement.

If an angle β is obtuse, its supplement S(β) is acute. Furthermore, the angle is less thanits supplement.

Reason. Assume α < R, where R denotes a right angle. From the comparison ofsupplements done in Proposition 5.27, we conclude S(α) > S(R). But by the definitionof a right angle R ∼= S(R). Hence S(α) > S(R) ∼= R, and because comparison isfor congruence classes, we get S(α) > R. Hence α < R < S(α), and by transitivity(Problem 10.2), we conclude α < S(α).

Similarly, we explain that β > R implies S(β) < R and S(β) < β.

Proposition 5.28 (All right angles are congruent).

Question. How is a right angle defined?

Answer. A right angle is an angle congruent to its supplementary angle.

Proof in the conventional style. Let α be a right angle, and β = S(α) ∼= α be itscongruent supplement. Similarly, we consider a second pair of right angles α′ andβ′ = S(α′) ∼= α′ being its congruent supplement.

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The question is whether both right angles α and α′ are congruent to each other. ByProposition 5.26 all angles are comparable. Hence the two angles α and α′ satisfy justone of the following relations.

Either (i) α < α′ or (ii) α ∼= α′ or (iii) α′ < α.

We need to rule out cases (i) and (iii) by deriving a contradiction. I only need to explaincase (i), because (iii) is similar. Now assume α < α′ towards a contradiction. By thecomparison of supplements from Proposition 5.27, we get S(α) > S(α′). Because a rightangle is congruent to its supplement we get α ∼= S(α) > S(α′) ∼= α′. As explained inProposition 5.24, angle comparison holds for congruence classes. Hence we get α > α′.

Thus, from the assumption α < α′, we have derive α > α′. Transitivity would implyα < α which is impossible. (Axiom III.4 states α ∼= α.)

Thus case (i) leads to a contradiction. Similarly, case (iii) leads to a contradiction.The only remaining possibility is case (ii): Any two given right angles α and α′ arecongruent.

Proposition 5.29 (About sums of angles). The sum of angles is defined on equiv-alence classes of congruent angles. It satisfies the following properties:

Commutativity If α + β exists, then β + α exists and β + α ∼= α + β.

Associativity (α + β) + γ ∼= α + (β + γ) and existence of one side implies existenceof the other one.

Difference α < γ if and only if there exists β such that α + β ∼= γ.

Comparison of Sums 1 If α ∼= β and γ < δ and β + δ exists, then α + γ exists andα + γ < β + γ.

Comparison of Sums 2 If α < β and γ < δ and β + δ exists, then α + γ exists andα + γ < β + γ.

Proof. By Hilbert’s Theorem 15 and Proposition 5.23, the sum of angles is defined onequivalence classes of congruent angles. Now we check the items stated:

Commutativity: Let α ∼= ∠ABG and β ∼= ∠GBC where points G and C lie both onthe same side of line AB. With that setup, α+ β ∼= ∠ABC, which is assumed toexist. Since the order of the two rays of an angle is defined to be chosen arbitrarily,

∠ABC = ∠CBA ∼= ∠CBG+ ∠GBA ∼= β + α

hence the latter angle exists and β + α ∼= α + β.

Associativity: The proof is left to the reader.

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Difference: Let α ∼= ∠ABG and γ ∼= ∠ABC where points G and C lie both on the

same side of line AB. With that setup, α < γ iff the ray−−→BG lies inside the angle

∠ABC iff α + β ∼= γ with β ∼= ∠GBC.

Comparison of Sums 1: Assuming that α ∼= β and γ < δ and β+ δ exists, we knowthere exists ε such that γ + ε ∼= δ. Hence

(α + γ) + ε ∼= α + (γ + ε) ∼= β + δ

and α + γ < β + γ where the former is shown to exist.

Comparison of Sums 2: Assuming α < β and γ < δ, we know there exist angles ηand ε such that α + η ∼= β and γ + ε ∼= δ. Hence

(α + γ) + η + ε ∼= (α + η) + (γ + ε) ∼= β + δ

and α + γ < β + γ where the former is shown to exist.

Proposition 5.30 (The Hypothenuse Leg Theorem). Two right triangles for whichthe two hypothenuse, and one pair of legs are congruent, are congruent.

Question. Of which one of SAS, SSA, SAA, ASA, SSS congruence contains the hypothenuse-leg theorem as a special case? Is there a corresponding unrestricted congruence theorem?

Answer. The hypothenuse-leg theorem is a special case of SSA congruence. There is nounrestricted SSA congruence theorem.

First proof of the hypothenuse-leg theorem 5.30 . Given are two right triangles �ABCand�A′B′C ′. As usual, we put the right angles at vertices C and C ′. We assume congruenceof the hypothenuses AB ∼= A′B′ and of one pair of legs AC ∼= A′C ′. We can build a kiteout of two copies of triangle �ABC and two copies of triangle �A′B′C ′.

To this end, one transfers angle ∠C ′A′B′ onto the ray−→AC into the opposite half

plane and gets ∠C ′A′B′ ∼= ∠CAB′′. Point B′′ can be chosen on the newly produced raysuch that A′B′ ∼= AB′′. By SAS-congruence it is easy to confirm �A′B′C ′ ∼= �AB′′Cand especially

(5.3) B′C ′ ∼= B′′C

Too, the two supplementary right angles at vertex C imply that points B,C and B′′ lieon a line. (Why?)

Next we transfer segment CA onto the ray opposite to−→CA and get point D such

that CA ∼= CD. We produce the two triangles in the right half of the figure. Hence SAScongruences using the right angle imply �ABC ∼= �DBC and �AB′′C ∼= �DB′′C.

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Since AB ∼= A′B′ ∼= AB′′ and DB ∼= AB ∼= AB′′ ∼= DB′′, one has constructed asymmetric kite with the four points

X := A, Y := D, Z1 := B, Z2 := B′′

as shown in the drawing. We can apply Hilbert’s kite-theorem 5.19 and conclude

Figure 5.30: Four right triangles yield a kite. When is it even a rhombus?

∠DAB = ∠XY Z1 ∼= ∠XY Z2 = ∠DAB′′

The congruence of the same angles is ∠CAB = ∠B′′AC. Hence SAS congruence impliesnow �CAB ∼= �B′′AC and

BC ∼= B′′C

Together with formula (5.3) above we conclude B′C ′ ∼= BC. Now a final SAS congruenceusing the right angles implies implies congruence �ABC ∼= �A′B′C ′ of the originallygiven triangles, as to be shown.

5.6 Constructions with Hilbert tools

Now we show that a right angle actually exists, and do some further basic constructionswith the Hilbert tools. The only means of construction are those granted by Hilbert’saxioms.

Definition 5.9 (Hilbert tools). By incidence axiom I.1 and I.2, we can draw a uniqueline between any two given points. By axiom III.1 and Proposition 5.2 from the very

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beginning, we transfer a given segment uniquely to a given ray. Finally by axiomIII.4, we transfer a given angle uniquely on a given ray into the specified half plane.Constructions done using only these means are called constructions by Hilbert tools.

Problem 5.3 (Drop a Perpendicular). Given is a line OA and a point B not onthis line. We have to drop the perpendicular from point B onto line OA.

Construction 5.1. Draw ray−−→OB. Transfer angle ∠AOB, into the half plane opposite

to B, with ray−→OA as one side. On the newly produced ray, transfer segment OB to

produce a new segment OC ∼= OB. The line BC is the perpendicular, dropped frompoint B onto line OA.

Figure 5.31: Drop the perpendicular, the two cases

Proof of validity. The line OA and the segment BC intersect, because B and C lie ondifferent sides of OA. I call the intersection point M . It can happen that O = M . Inthat special case A �= M and

(**) ∠AMB ∼= ∠AMC

But these is a pair of congruent supplementary angles, because M lies between B andC. By definition, an angle congruent to its supplementary angle is a right angle. Hence∠AMB is a right angle.

In the generic situation O �= M , we distinguish two cases

(i) On the given line OA, points M and A lie on the same side of O.

(ii) Point O lies between M and A.

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In both cases we show by the SAS-congruence theorem that the two triangles �OMBand �OMC are congruent. Indeed, the angles at vertex O are congruent, both in case(i) and (ii):

In case (i) , the rays−→OA =

−−→OM are equal, and ∠MOB = ∠AOB ∼= ∠AOC = ∠MOC

follows from the construction.

In case (ii) , the rays−→OA and

−−→OM are opposite. Thus ∠MOB and ∠AOB, as well as

∠MOC and ∠AOC are supplementary angles. Again ∠AOB ∼= ∠AOC becauseof the angle transfer done in the construction. Now Theorem 14 of Hilbert (seeProposition 5.14 above) tells that supplements of congruent angles are congruent.Hence we get ∠MOB ∼= ∠MOC once again.

The adjacent sides OB ∼= OC are congruent by construction, too. Finally the commonsides OM is congruent to itself. The drawing stresses the pieces matched to prove thecongruence. Because the two triangles are congruent, the corresponding angles at vertex

Figure 5.32: Congruences needed in the two cases

M are congruent, too. Hence

(*) ∠OMB ∼= ∠OMC

which is a pair of congruent supplementary angles. A right angle is, by definition, anangle congruent to its supplementary angle . Hence either ∠OMB is a right anglebecause of formula (*), or, in the special case O = M , angle ∠AMB is a right anglebecause of (**).

Proposition 5.31 (Supplements of acute and obtuse angles). Let R denote aright angle. For any angle γ and its supplement S(γ) exactly one of the following threecases occurs: Either (1) or (2) or (3).

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(1) γ < R, S(γ) > R and γ < S(γ)

(2) γ ∼= R, S(γ) ∼= R and γ ∼= S(γ)

(3) γ > R, S(γ) < R and γ > S(γ).

Proof. All angles are comparable (see Proposition 5.26). For angle γ and the right angleR, exactly one of the three cases holds:

Either (i) γ < R or (ii) γ ∼= R or (iii) γ > R.

From SAS congruence (Theorem 12 of Hilbert, see Proposition 5.10 above), we know thatα ∼= β implies S(α) ∼= S(β). From Proposition 5.27 about comparison of supplements,we know that α < β implies S(α) > S(β). Hence with the help of Propositions 5.23and 5.24 we get:

in case (i), one concludes S(γ) > S(R) ∼= R > γ, and hence (a) γ < S(γ),

in case (ii), one concludes S(γ) ∼= S(R) ∼= R ∼= γ, and hence (b) γ ∼= S(γ),

in case (iii), one concludes S(γ) < S(R) ∼= R < γ, and hence (c) γ > S(γ).

By Proposition 5.26 above, all angles are comparable. Hence the two angles γ and itssupplement S(γ) satisfy either (a) or (b) or (c). This observation allows one to get theconverse of the conclusions shown above. For example, (a) excludes both (ii) and (iii),and hence implies (i). Thus we get

(a) implies (i) , (b) implies (ii) , (c) implies (iii); and finally(a) if and only if (i) , (b) if and only if (ii) , (c) if and only if (iii).

This leads to the mutually exclusive cases (1) (2) (3), as originally stated.

Corollary 7 (All right angles are congruent).

Proof. The proposition 5.31 about supplements of acute and obtuse angles yields aneasy proof that all right angles are congruent:

Let R be the right angle as has been constructed in Construction 5.1. Now supposeγ = R′ is another right angle. By definition of a right angle, this means that γ ∼= S(γ).Hence case (b) above occurs, which implies (ii): γ ∼= R. You see that existence of aright angle make proving its uniqueness a bit easier.

Proposition 5.32 (Converse Isosceles Triangle Proposition). [Euclid I.6, Theo-rem 24 of Hilbert] A triangle with two congruent angles is isosceles.

Question. Formulate the theorem with specific quantities from a triangle �ABC.

Answer. If α ∼= β, then a ∼= b.

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Proof. Given is a triangle �ABC with α ∼= β. By Hilbert’s Theorem 19 (see Proposi-tion 5.22 above) and Proposition 5.23, congruence of angles is an equivalence relation.Hence α ∼= β implies β ∼= α. In the preliminary version given as Proposition 5.13, wehave shown that α ∼= β and β ∼= α together imply a ∼= b. Hence the �ABC is isosceles,as to be shown.

Proposition 5.33 (Existence of an Isosceles Triangle). For any given segmentAB, on a given side of line AB, there exists an isosceles triangle.

Of course, this triangle is not unique.

Construction 5.2 (Construction of an isosceles triangle). Given is a segmentAB. Choose any point P not on line AB, in the half plane specified. Compare the twoangles ∠BAP and ∠ABP . In the drawing, ∠BAP is the smaller angle. Transfer the

smaller angle, with ray−→BA as one side, and the newly produced ray rB in the same half

plane as P . Ray rB and segment AP do intersect. The intersection point C lies in thehalf plane as required, and �ABC is isosceles.

Figure 5.33: Construction of an isosceles triangle

Proof of validity. By Proposition 5.26, all angles are comparable. Comparison of thetwo angles ∠BAP and ∠ABP leads to one of the three possibilities:

Either (i) The two angles are congruent

or (ii) α = ∠BAP < ∠ABP = β

or (iii) α = ∠BAP > ∠ABP = β.

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It is enough to consider cases (i) and (ii). In case (i) let C = P . In case (ii) we have

transferred the smaller angle ∠BAP , onto ray−→BA, and the newly produced ray rB in

the same half plane as P . By the Crossbar Theorem, a segment going from one sideof an angle to the other, and a ray in the interior of an angle always intersect. By theCrossbar Theorem, ray rB and segment AP intersect. The intersection point C liesbetween A and P , hence both points C and P lie in the same half plane of line AB.Hence point C lies in the half plane as required.

By construction, the base angles of triangle�ABC are congruent. Hence the triangleis isosceles by Euclid I.6 (Converse Isosceles Triangles).

Remark. One may suggest to make it part of the construction how to compare the twoangles ∠BAP and ∠ABP . One possibility is to transfer both angles to the other vertex,B or A, respectively. Only the newly produced ray from the smaller angle does intersectthe opposite side of �ABP .

Problem 5.4 (Erect a Perpendicular). Given is a line l and a point R on this line.We have to erect the perpendicular at point R onto line l.

Construction 5.3. Choose any point A �= R on line l. Transfer segment AR onto

the ray opposite to−→RA to get a segment RB ∼= AR. As explained in Construction 5.2,

construct any isosceles triangle �ABC. The line CR is the perpendicular to the givenline l through point R.

Figure 5.34: Erect a perpendicular

Reason for validity. The two triangles �RAC and �RBC are congruent by SAS con-gruence. Indeed, ∠RAC ∼= ∠RBC by the construction of the isosceles triangle. Fur-thermore, we have a pair of congruent adjacent sides: AR ∼= BR by the constructionabove, and AC ∼= BC because �ABC is isosceles.

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Now the triangle congruence �RAC ∼= �RBC implies that ∠ARC ∼= ∠BRC. Butthese two angles are supplementary angles. Because congruent supplementary anglesare right angles, we have confirmed that ∠ARC is a right angle, as to be shown.

Definition 5.10 (The perpendicular bisector). The line through the midpoint of asegment and perpendicular to the segment is called the perpendicular bisector.

Problem 5.5. Given is any segment AB. Construct the perpendicular bisector.

Construction 5.4. Construct two different isosceles triangles over segment AB. Theline connecting the third vertices of the two isosceles triangles is the perpendicular bi-sector.

Figure 5.35: The perpendicular bisector via the kite

Proof of validity. Let �ABC be the first isosceles triangle, and �ABD be the secondone. We want to proceed as in Hilbert’s theorem 17 (see Proposition 5.19 above).

Question. Why do the points A and B lie on different sides of CD? We need just touse the Lemma 5.3 to SSS-congruence!

Answer. If points A and B would lie on the same side of line CD, the congruence�ACD ∼= �BCD would imply A = B, contradicting to the endpoints of a segmentbeing different.

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Because A and B lie on different sides of line CD, the segment AB intersects theline CD, say at point M . Now we have the symmetric kite ACBD with two congruenttriangles (left and right in the figure): �ACD ∼= �BCD.

Question. For convenience, please repeat how this follows just as in Theorem 17 (seeProposition 5.19 above).

Answer. Addition or subtraction of the base angles of the two isosceles triangles �ABCand �ABD yields the congruent angles ∠CAD ∼= ∠CBD—either using sums or dif-ferences of congruent base angles. The triangles �ACD and �BCD have congruentangles at A and B, and two pairs of congruent adjacent sides AC ∼= BC and AD ∼= BD.Now SAS congruence implies �ACD ∼= �BCD.

Of the three points M,C,D exactly one lies between the two others. We can assumethat C does not lie between D and M , this assumption can be achieved by possiblyinterchanging the names C and D. Next we show that �ACM ∼= �BCM . Indeed,

Figure 5.36: Get the perpendicular bisector via: (a) a convex kite, (b) a nonconvex kite

the triangles �ACM and �BCM have congruent angles at the common vertex C, andtwo pairs of congruent adjacent sides AC ∼= BC and MC ∼= MC. Now we get by SAScongruence that �ACM ∼= �BCM . Matching pieces are stressed in the drawing.

From the last triangle congruence we get ∠AMC ∼= ∠BMC. BecauseM lies betweenA and B, these are two congruent supplementary angles. Hence they are right angles.Too, the triangle congruence implies AM ∼= BM . Hence M is the midpoint of segmentAB. We have shown that the segment AB and the line CD intersect perpendicularlyat the midpoint of segment AB. Hence CD is the perpendicular bisector.

Remark. There are the possibilities of an (a) convex kite, or (b) a non convex kite.

(a): If the two isosceles �ABC and �ABD lie on different sides of AB, points C andD are on different sides of AB, and it is clear that the segment CD intersects theline AB.

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(b): Too, it is possible to use two different isosceles triangles on the same side of AB.To get the second isosceles triangle �ABD, one transfers as base angles anytwo congruent angles which are less than the base angles of the first triangle�ABC. The figure ACBD is still a symmetric—but non convex—kite. As provedin Hilbert’s Theorem 17 (see Proposition 5.19 above), the two triangles left andright in the figure are still congruent: �ACD ∼= �BCD. The line CD stillintersects the segment AB, because points A and B are on different sides of CD,but the two segments AB and CD do not intersect each other.

Construction 5.5 (Solution— the way one really wants it). Construct any isoscelestriangle over the given segment. Drop the perpendicular from its third vertex.

Figure 5.37: Construction of the perpendicular bisector—the natural way

Independent proof of validity. Let �ABC be the isosceles triangle constructed at thefirst step. In the next construction step, the base angle ∠BAC is reproduced along the

ray−→AB to the other side of line AB, opposite to point C. Finally, one transfers segment

AC onto the newly produced ray, and gets the congruent segment AD ∼= AC.

Question. Show the congruence

(up-down) �ABC ∼= �ABDAnswer. This follows by SAS congruence. Indeed, the two triangles have the commonside AB, the two sides AC ∼= AD are congruent, and the angles ∠BAC and ∠BAD arecongruent, both by construction.

Hence the construction has produced two isosceles triangles with base AB, whichare congruent to each other. Next we show that the triangles

(left-right) �ACD ∼= �BCDare congruent.

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Figure 5.38: (up-down) �ABC congruent �ABD (left-right) �ACD congruent �BCD(upleft-upright) �ACM congruent �MCB

Question. Explain how this congruence is shown.

Answer. This follows by SAS congruence. All four sides AD ∼= AC ∼= BC ∼= BD arecongruent. Furthermore, the two triangles have congruent angles ∠CAD ∼= ∠CBD, asfollows from the congruences ∠CAB ∼= ∠DAB, and ∠ABC ∼= ∠ABD of base angles ofthe two isosceles triangles shown in (up-down)—and angle addition.

The line AB and the segment CD intersect, because C and D lie on different sidesof AB. I call the intersection point M . Finally we show that the triangles

(upleft-upright) �ACM ∼= �BCMare congruent.

Question. Explain how this congruence is shown.

Answer. This follows by SAS congruence. Indeed, the two triangles have the commonside CM , the two sides AC ∼= AB are congruent, and the angles ∠ACM and ∠BCMare congruent because of congruence (left-right).

Hence the segments AM ∼= BM are congruent— line CD bisects the segment AB.

Question. Why does M lies between A and B.

Answer. We know that the segments AM ∼= BM are congruent—this is impossible fora point on the line AB outside the segment AB.

Too, the angles ∠AMC ∼= ∠BMC are congruent supplementary angles, and henceright angles. Thus we have confirmed that CD is the perpendicular bisector of segmentAB, as to be shown.

Another attempt to prove validity, it does not yet work! As already shown in the con-struction for dropping a perpendicular, the line CD is perpendicular to the given lineAB. Let M be the intersection point of these two lines.

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We need still to show that M is the midpoint of segment AB. To this end, we provethat the triangles �AMC ∼= �BMC are congruent. This can be shown only usingSAA-congruence from Proposition 5.45 below!

By the converse isosceles triangle Proposition, we know that AC ∼= BC. Becauseof congruence of vertical angles, we know that ∠BMC ∼= ∠AMD, the latter angle wasalready shown to be a right angle. Hence, all put together, we get ∠BMC ∼= ∠AMD ∼=R ∼= ∠AMC as expected. Finally, we have the congruence base angles ∠BAC ∼= ∠ABCby construction.

Hence via SAA-congruence from Proposition 5.45 below, we look forward to es-tablishing the triangle congruence �AMC ∼= �BMC, and hence confirming AM ∼=MB.

Question. Why does this proof not work, at this point of the development?

Answer.

5.7 The exterior angle theorem and its consequences

Recall that an exterior angle of a triangle is the supplement of an interior angle.

Proposition 5.34 (The Exterior Angle Theorem). [Euclid I.16. Theorem 22 inHilbert] The exterior angle of a triangle is greater than both nonadjacent interior angles.

Proof. For the given �ABC, we can choose point D on the ray opposite to−→AB, such

that AD ∼= CB. We compare the two nonadjacent interior angles γ = ∠ACB andβ = ∠ABC to the exterior angle δ = ∠CAD. As a first step, we show

Lemma 5.4. The exterior angle δ = ∠CAD is not congruent to the interior angle∠ACB = γ.

Proof of Lemma‘5.4. Assume towards a contradiction that δ ∼= γ.The supplementary angle of δ is α = ∠CAB. The supplementary angle of γ is

∠ACE, with a point E on the ray opposite to−−→CB. Supplementary angles of congruent

angles are congruent (Theorem 14 in Hilbert, see Proposition 5.14 above). Hence

?1 ∠CAB ∼= ∠ACE

On the other hand, we use SAS-congruence for the two triangles �ABC and �A′B′C ′with

A′ := C , B′ := D , C ′ := A

Those two triangles would be congruent by SAS congruence. Indeed the angles atC and C ′ are congruent by our assumption γ = ∠ACB ∼= ∠CAD = δ. Too, theadjacent sides are pairwise congruent because of CB ∼= AD = C ′B′ by construction,

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Figure 5.39: The impossible situation of a congruent exterior angle

and CA ∼= AC = C ′A′. From Axiom III.5, we conclude that the corresponding anglesat A and A′ = C are congruent:

?2 ∠CAB ∼= ∠C ′A′B′ = ∠ACD

We have shown that angle ∠CAB is congruent to both (?1) ∠ACE and (?2) ∠ACD.Now we use the uniqueness of angle transfer. Transferring angle ∠CAB with one

side−→CA, into the half plane not containing B, yields as second side of the angle once

the ray−−→CE and the second time

−−→CD. Hence these two rays are the same.

Hence points E and D both lie on the line la := BC. In other words, the four pointsB,C,D,E all lie on one line. Because B and D lie on opposite rays with vertex A,these two points are different: B �= D. Note that just this simple remark is not true inelliptic geometry!

Because a line is uniquely specified by two of its points, this implies la = BC = BD.We have just seen that point C lies on this line. By construction, point A lies on theline BD, too. Hence all the points A,B,C,D,E lie on the same line. This contradictsthe definition of a triangle. By definition of a triangle, its three vertices do not lie onone line. This contradiction confirms the original claim δ � γ.

Remark. To get a reminiscence to point symmetry about the midpoint M of AC, we canchoose AB ∼= CE. With the choice AB ∼= CE, we get even E = D, but still D �= B.

Remark. In spherical geometry, the figure constructed does exist. We do not get acontradiction, the conclusion is just that the two points B and D are antipodes, andthe vertices A and C lie on two different lines BAD and BCD through these twoantipodes. Too, the sum of the segments is congruent going both ways from B to D.The triangles �ABC ∼= �A′B′C ′ = �CDA are congruent, by SAS congruence (givenin Proposition 5.10 above). Hence BC ∼= DA and CD ∼= AB, and hence the sumsBC + CD ∼= DA+ AB are congruent.

The next Lemma rules out the possibility δ < γ.

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Figure 5.40: Two ways from B to D

Lemma 5.5. The exterior angle δ = ∠CAD is not smaller that the interior angle∠ACB = γ.

Proof of Lemma 5.5. Suppose towards a contradiction that δ < γ. We transfer the

Figure 5.41: The impossible situation of a smaller exterior angle

exterior angle δ = ∠CAD with one side−→CA, into the half plane containing B. The

second side of the angle we get is a ray−−→CB′ inside the angle ∠ACB. By the Crossbar

Theorem, it meets the segment AB in a point B′.We can now apply the first Lemma 5.4 to �AB′C. This smaller triangle would have

exterior angle δ congruent to the interior angle γ′ = ∠ACB′. This is impossible byLemma 5.4.

Thus we have both ruled out the possibility that δ = γ, or that δ < γ. By Propo-sition 5.26, all angles are comparable. Hence there remains only the possibility thatδ > γ.

Lemma 5.6. The exterior angle δ = ∠CAD is greater than the interior angleβ = ∠ABC.

Proof of Lemma 5.6. We compare angle β = ∠ABC to the exterior angle δ = ∠CAD.

Choose any point F on the ray opposite to−→AC. The vertical angles δ = ∠CAD ∼=

∠FAB = δ′ are congruent by Euclid I.15. Now we are back to the case already covered,because the interior angle β = ∠ABC and exterior angle ∠FAB lie on opposite sidesof triangle side AB, which is also part of one side of each of these angles. By Lemma5.4and 5.5, we conclude ∠FAB > ∠ABC and hence δ > β.

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Figure 5.42: Comparing the other nonadjacent angle

Remark. Here is an other way to prove Lemma 5.6: Define �A3B3C3 by setting A3 :=A, B3 := C, C3 := B. Now we can apply Lemma5.4 and 5.5 to this new triangle andget δ′ = δ3 > γ3 = β. By congruence of the vertical angles δ = ∠CAD ∼= ∠FAB = δ′

we get δ > β.

Thus the proof of the exterior angle theorem is finished.

Proposition 5.35. [Euclid I.17.] The sum of any two interior angles of a triangle isless than two right angles.

Proposition 5.36 (Immediate consequences of the exterior angle theorem).

(i) Every triangle can have at most one right or obtuse angle.

(ii) The base angles of an isosceles triangle are acute.

(iii) The foot point of a perpendicular is unique.

(iv) Given a line l and a point O not on l. At most two points of l have the samedistance from O.

(v) A circle and a line can intersect in at most two points.

Proof. (i): Suppose angle α of �ABC is right or obtuse: α ≥ R. By Proposition 5.31,its supplement is right or acute: S(α) ≤ R. But the supplement is an exteriorangle: δ = S(α). By the exterior angle theorem, the two other nonadjacentinterior angles of the triangle are less than that exterior angle. Hence they areacute: β, γ < δ = S(α) ≤ R, and hence β, γ < R.

(ii): The two base angles being congruent, by (i), they cannot be both right or obtuse.

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(iii): Given a line l and a point O not on l. Suppose there are two perpendiculars, withfoot points F and G. The �OFG would have two right angles, contradicting (i).

(iv): Given a line l and a point O not on l. Suppose towards a contradiction there

Figure 5.43: No three points have the same distance from a line

are three points A,B,C on the line l such that OA ∼= OB ∼= OC. Of any threepoints, one lies between the two others (Theorem 4 of Hilbert). We may supposeby renaming that A ∗B ∗ C.

The base angle α of triangle �OBC is an exterior angle of the other triangle�OAB. Hence the exterior angle theorem yields γ > α. But the role of the twotriangles can be switched: The base angle γ of triangle �OAB is an exterior angleof the other triangle�OBC. Hence the exterior angle theorem yields α < γ. Bothinequalities together are impossible. This contradiction implies there can exist atmost two points on a line which have congruent distances from point O.

(v) For the case that the center O of the circle does not lie on the line, the statementis confirmed by item (iv) above.

For the case that the center O of the circle lies on the line, the statement followsfrom line separation and the uniqueness of segment transfer stated in Proposi-tion 5.2.

Definition 5.11 (Alternate interior angles or z-angles). Let a transversal t inter-sect two lines a and b at the points A and B. A pair of alternate interior angles or simplyz-angles are two angles with vertices A and B, lying on different sides of the transversal.They have as one pair of sides lies on the transversal and contains the segment AB, theremaining two sides are rays lying on a and b.

Proposition 5.37 (Congruent z-angles imply parallels). [Euclid I.27] If two linesform congruent z-angles with a transversal, they are parallel.

Proof. This is an immediate consequence of the exterior angle theorem. We argue bycontradiction. Suppose the two lines would intersect in point C. One of the z-angles

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is an exterior angle of triangle �ABC, the other one is a nonadjacent interior angle.Lemma 5.4 does imply that the two z-angles are not congruent— contradicting theassumption from above. Hence the two lines a and b cannot intersect.

Figure 5.44: A pair of z-angles.

Remark. Referring to the pair of z-angles shown in the figure on page 178 the Proposi-tion 5.37 tells that

α ∼= β ⇒ a ‖ bThis is equivalent to its contrapositive

a ∦ b⇒ α �∼= β

which relates even more directly to the exterior angle theorem.

As a consequence of Euclid I.27, the existence of a parallel can be proved in neutralgeometry. One parallel to l through point P is conveniently constructed as ”doubleperpendicular”.

Proposition 5.38 (Existence of a parallel). For every line l and for every point Plying not on l, there exists at least one parallel m to l through point P .

Proof. Given is line l and a point P not on l. One drops the perpendicular from point Ponto line l and denotes the foot point by F . Next, one erects at point P the perpendicularto line PF . Thus, one gets a line, which we call m. Because the two lines l and m form

congruent z-angles with the transversal←→PF , Euclid I.27 or I.28 imply that l and m are

parallel. This leaves open the question whether or not m is the unique parallel to line lthrough point P .

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Recall that the Euclidean Parallel Postulate 2.2 postulates both existence and unique-ness of the parallel to a given line through a given point. On the other hand, Hilbert’sParallel Postulate for plane geometry 2.3 postulates only the uniqueness of the parallelto a given line through a given point. Because of Proposition 5.38, Euclid’s and Hilbert’sparallel postulate turn out to be equivalent for any Hilbert plane. In other word, wecan state the definition equivalent to definition 1.2:

Definition 5.12 (Pythagorean plane). A Pythagorean plane is a Hilbert plane forwhich the Euclidean parallel postulate holds.

Problem 5.6. Given is a convex quadrilateral with two pairs of congruent oppositesides. Since the quadrilateral is convex, each diagonal separates the quadrilateral intotwo triangles in opposite half planes of their common side. Prove in neutral geometrythat the two pairs of opposite sides are parallel. Provide a drawing.

Figure 5.45: A parallelogram in neutral geometry.

Answer. Given is the quadrilateral �ACBD with two pairs of congruent opposite sidesAC ∼= BD and BC ∼= AD. We draw the diagonal AB and get the congruent triangles�ABC ∼= �BAD, as one confirms by SSS congruence. Hence the angles β = ∠ABCand α = ∠BAD are congruent. We see that the diagonal AB transverses the lines ofthe opposite sides DA and BC with congruent z-angles. By Euclid I.27, we concludethat the opposite sides are parallel.

Remark. Similar to the exterior angle theorem, these are facts of neutral geometry. Theyare is valid both in Euclidean and hyperbolic geometry. Note that Euclid I.2 throughI.28 are theorems that hold for every Hilbert plane. Indeed Euclid I.29 is the firsttheorem in Euclid’s elements that uses the Euclidean parallel postulate.

Warning. The converse of Euclid I.27 is does not hold in hyperbolic geometry. Indeed,in hyperbolic geometry, parallels can form non-congruent z-angles with a transversal.

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Only in Euclidean geometry is it true that a ‖ b ⇒ α ∼= β. In more commonlanguage:

Only in Euclidean geometry can parallel lines be used as a means

for the transport of an angle.

Problem 5.7 (Addendum the the extended ASA-Congruence Theorem). Inthe situation of the extended ASA-congruence theorem, what happens in case that A′B′ <AB or A′B′ > AB?

Question. Using the exterior angle theorem and Pasch’s axiom, get some results for thefirst case.

Answer. In the case A′B′ < AB, one can conclude that the two rays r′A and r′B still dointersect.

Detailed proof. I show that the two rays do intersect. One transfers segment AB to the

ray−−→A′B′ and gets a segment A′B2 ∼= AB with A′ ∗B′ ∗B2.We apply the extended ASA theorem to�ABC and segment A′B2, and get a triangle

�A′B2C2 ∼= �ABC.Now apply Pasch’s axiom to triangle �AB2C2 and line l on the ray newly produced

r′B. This line intersects the triangle side A′B2 in B′, by construction. Hence Pasch’saxiom tells that line l intersects a second side of �AB2C2, too, or goes through pointC2.

Line l does not go through point C2. Otherwise, one would get a contradiction tothe exterior angle theorem.

Question. For which triangle do you get a this contradiction?

Answer. �B′B2C would have both the interior angle β at vertex B2, and the exteriorangle β at vertex B′.

Line l does not intersect line B2C2. Again, one would get a contradiction to theexterior angle theorem. Alternatively, one can say that the two lines l and B′C ′ areparallel, because they form congruent z-angles with line A′B′.

Hence Pasch’s axiom implies that line l intersects the third side of �AB2C2, whichis segment A′C2. We call the intersection point C ′. Thus we have produced a �A′B′C ′,the angles of which at vertices A′ and B′ are congruent to the corresponding angles of�ABC.

Remark. In Euclidean geometry, the two rays r′A and r′B always intersect, no matterwhether A′B′ < AB, A′B′ ∼= AB, or A′B′ > AB. In all three cases, the �A′B′C ′ issimilar to �ABC.

Remark. Here is what happens in hyperbolic geometry: In hyperbolic geometry, similartriangles are always congruent, this implies that �A′B′C ′ and �ABC are not similarif either A′B′ < AB or A′B′ > AB!

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Figure 5.46: The extended ASA congruence once more

In the case just considered, A′B′ < AB and γ′ > γ. In the opposite case thatA′B′ > AB, the second triangle �A′B′C ′ has either a different angle γ′ < γ, or doesnot exist at all. The last possibilities occurs because the two rays r′A and r′B do notintersect at all. This happens always, once the segment A′B′ is long enough.

Proposition 5.39 (Comparison of sides implies comparison of angles). [EuclidI.18, Theorem 23 of Hilbert] In any triangle, across the longer side lies the greater angle.

Proof. In �ABC, we assume for sides AB and BC that c = AB > BC = a. The issueis to compare the angles α = ∠CAB and γ = ∠ACB across these two sides.

We transfer the shorter side BC at the common vertex B onto the longer side. Thusone gets a segment BD ∼= BC, with point D between B and A. Because the �BCD isisosceles, it has two congruent base angles

δ = ∠CDB ∼= ∠DCB

Because B ∗D ∗ A, we get by angle comparison at vertex C

δ = ∠DCB < γ = ∠ACB

Now we use the exterior angle theorem for �ACD. Hence

α = ∠CAB < δ = ∠CDB

By transitivity, these three equations together imply that α < γ. Hence the angle αacross the smaller side CB is smaller than the angle γ lying across the greater side AB.In short, we have shown that c > a⇒ γ > α.

Proposition 5.40 (Comparison of angles implies comparison of sides). [EuclidI.19] In any triangle, across the greater angle lies the longer side.

Proof. In �ABC, we assume for two angles that α = ∠CAB < γ = ∠ACB. The issueis to compare the two sides BC and AB lying across the angles and show a = BC <AB = c. As shown in Proposition 5.6, any two segments are comparable.

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Figure 5.47: Across the longer side lies the greater angle

Question. Please repeat the reasoning for convenience.

Answer. We transfer segment BC along the ray−→BA, and get a segment BD ∼= BC. By

Theorem 4 in Hilbert, of the three points A,B,D on a line AB, exactly one lies betweenthe two others. This leads to the following three cases: Either

(i) A ∗D ∗B and a < c. or (ii) A = D and a = c. or (i) D ∗ A ∗B and a > c.

We can now rule out cases (ii) and (iii).

In case (ii), the �ABC is isosceles, and Euclid I.5 would imply α = γ, contrary tothe hypothesis about these two angles.

In case (iii), Euclid I.18 or Theorem 23 of Hilbert, would imply α > γ, contrary tothe hypothesis.

Hence only case (i) is left, and a < c, as to be shown. In short, we have shown thatγ > α⇒ c > a.

Question. Explain how Euclid I.18 and Euclid I.19 are logically related.

Answer. Euclid I.18 in shorthand: c > a⇒ γ > α.

Euclid I.19 in shorthand: γ > α⇒ c > a.

Euclid I.19 is the converse of Euclid I.18.

Question. Does Euclid I.19 follow from Euclid I.18 by pure logic? Why not?

Answer. No, the converse does not follow purely by logic.

Question. Which fact does the proof above work nevertheless?

Answer. Because any two segments are comparable, we get the converse, nevertheless.

Corollary 8. A triangle with two congruent angles is isosceles.

Proof. Assume α ∼= γ for the given triangle. Since c > a⇒ γ > α and c < a⇒ γ < α,and any two segments are comparable, only a ∼= c is possible.

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Proposition 5.41. The hypothenuse is the longest side of a right triangle.

Proof. By Proposition 5.36(i), a triangle can have at most one right or obtuse angle.Hence a right triangle �ABC with the right angle at vertex C has acute angles at thevertices A and B. As shown in Proposition 5.40 (Euclid I.19), comparison of angles ofa triangle implies comparison of its sides. Hence the two legs across of the acute anglesA or B are shorter than the hypothenuse across the right angle.

Proposition 5.42 (The foot point has the shortest distance). Given is a line anda point O not on the line. The foot point F of a perpendicular from O to the line is theunique among all points of the line, which has the shortest distance.

Figure 5.48: The foot point has the shortest distance

Proof. Call the given line l and let O be the given point not on l. Take any point A �= Fon the line l. The �AFO has a right angle at vertex F . The hypothenuse AO acrossto F is the longest side.

Remark. Because all other point on l except the foot point have strictly greater distancefrom O, we can once more conclude that the foot point of the perpendicular is unique.

Proposition 5.43 (The Triangle Inequality). [Euclid I.20] In any triangle, the sumof two sides is greater than the third side.

Proof. In �ABC, we compare the sum BA + AC to the third side BC. According toDefinition 5.2 the sum BA+ AC is obtained by transfer of the segment AC to the ray

opposite to−→AB. We get segment DA ∼= AC and an isosceles �DAC. Its two congruent

base angles areδ = ∠CDA ∼= ∠DCA

Because point A lies between D and B, angle comparison at vertex C yields

δ ∼= ∠DCA < ∠DCB = η

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Figure 5.49: The triangle inequality

Now we use Euclid I.19 for the larger �DCB. Hence the side BC, across the smallerangle δ is smaller than the side DB lying across the greater angle η: DB > BC. Forthe original �ABC, this shows that indeed CA+ AB > CB.

Corollary 9. For any three points A,B,C we have the inequality

AC ≤ AB +BC

and AC ∼= AB + BC holds if and only if point B lies between A and C or is equal toone of them.

Proposition 5.44. The sum of the segments of any polygon is equal to or longer thanthe distance of its first and last vertex.

The sum of the polygon segments A0A1, A1A2, A2A3, . . . , An−1An is equal to the dis-tance of its first vertex A0 to its last vertex An if and only if all vertices lie between themand occur in the order A0 ∗ A1 ∗ A2 ∗ A3 ∗ · · · ∗ An.Proof. We proceed by induction on the number of segments of the polygonA0A1, A1A2, A2A3, . . . , An−1An. The assertion holds for a polygon A0A1, A1A2 withtwo segments since A0A2 ≤ A0A1 + A1A2 according to the triangle inequality, and itscorollary.

For the induction step, we proceed from n to n+ 1 segments. We know that

A0An+1 ≤ A0An + AnAn+1

by the triangle inequality and

A0An ≤ A0A1 + A1A2 + · · ·+ An−1Anby the induction assumption. The associativity of the segment addition has alreadybeen stated as Proposition 5.7. Hence

A0An+1 ≤ A0An + AnAn+1 ≤ (A0A1 + A1A2 + · · ·+ An−1An) + AnAn+1∼= A0A1 + A1A2 + · · ·+ An−1An + AnAn+1

as to be shown.

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Proposition 5.45 (SAA-Congruence Theorem). [Theorem 25 in Hilbert] Assumetwo triangles have a pair of congruent sides, one pair of congruent angles across thesesides, and a second pair of congruent angles adjacent to these sides. Then the twotriangles are congruent.

Figure 5.50: SAA congruence

Proof. Let the triangles be�ABC and�A′B′C ′, and assume thatAB ∼= A′B′, ∠BAC ∼=∠B′A′C ′ and ∠ACB ∼= ∠A′C ′B′. We choose a point C ′′ on the ray

−−→A′C ′ such that

AC ∼= A′C ′′. By axiom III.5

γ = ∠ACB ∼= ∠A′C ′′B′

Indeed, the SAS congruence even implies

(*) �ABC ∼= �A′B′C ′′

On the other hand, by assumption

γ = ∠ACB ∼= ∠A′C ′B′

If C ′ �= C ′′, an exterior angle of �C ′C ′′B′ would be congruent to a nonadjacent interiorangle, which is impossible by the exterior angle theorem. Hence we can conclude thatC ′ = C ′′. Because of (*), this implies �ABC ∼= �A′B′C ′, as to be shown.

Question. How does this theorem differ from the ASA congruence (Hilbert’s Theorem13, and Proposition 5.12 above)?

Answer. Of the two pairs of angles that are given (or compared), one pair lies acrossthe pair of given sides.

Second proof of the hypothenuse-leg theorem 5.30. Take two right�ABC and�A′B′C ′with b = b′, c = c′ and γ = γ′ = R. One transfers segment CB and segment C ′B′, both

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Figure 5.51: The hypothenuse leg theorem

on the ray opposite to−−→C ′B′, and get new segments C ′D0 ∼= CB as well as C ′E ′ ∼= C ′B′.

From the construction and SAS congruence, we conclude

(1) �ABC ∼= �A′D0C , �A′B′C ′ ∼= �A′E ′C ′

Hence especially

(2) A′B′ ∼= A′D0 ∼= A′E ′

By Proposition 5.36(iv), at most two points can have the same distance from a line.Hence not all three pointsB′, D0, E ′ can be different. The only possibility left isD0 = B′,because they both lie on the opposite side of A′C ′ than B′.

Question. For completeness, explain once more. Assume that D0 �= E ′ towards a con-tradiction. Take the case that B′ ∗D0 ∗E ′ shown in the drawing. (The other cases canbe dealt with similarly.)

Independent answer. There are two isosceles �B′A′D0 and �DA′E ′.Question. What would happen with their base angles at vertex D0?

Answer. The two base angles would be supplementary.

Question. Why is this impossible?

Answer. This is impossible, because the exterior angle theorem would imply that eachone of them is larger than the other one.

This contradiction leave only the possibility that D0 = E ′.

Now the required congruence �ABC ∼= �A′B′C ′ follows from D0 = E ′ and (1).

Definition 5.13 (The angular bisector). The ray in the interior of an angle whichbisects the angle into two congruent angles is called the angular bisector.

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Proposition 5.46 (Uniqueness). The angular bisector is unique.

Proof. Suppose both rays−−→BG and

−−→BG′ bisect angle ∠BAC. The angles α := ∠ABG and

α′ := ∠ABG′ are comparable, because all angles are comparable by Proposition 5.26.Suppose towards a contradiction that α < α′. As specified in Proposition 5.29, one

can add inequalities of angles. Hence we conclude that

(?) ∠ABC = ∠ABG+ ∠GBC ∼= α + α < α′ + α′ ∼= ∠ABG′ + ∠G′BC ∼= ∠ABC

which is impossible. Similarly, the case α > α′ can be ruled out. Hence α ∼= α′, and

hence by uniqueness of angle transfer−−→BG =

−−→BG′, as to be shown.

Problem 5.8. Given is any angle ∠BAC. Construct the angular bisector.

Construction 5.6. We choose the segments on its sides to be congruent, thus assuming

AB ∼= AC. Draw the line BC, and transfer the base angle ∠ABC to the ray−−→BC, on

the side of line BC opposite to vertex A. On the new ray, we transfer segment AB to

get the new segment BD ∼= BA. The ray−−→AD is the bisector of the given angle ∠BAC.

Figure 5.52: The angular bisector

Question. Reformulate the description of this construction precisely, and as short aspossible.

Answer. One transfers two congruent segments AB and AC onto the two sides of theangle, both starting from the vertex A of the angle. The perpendicular, dropped fromthe vertex A onto the segment BC, is the angular bisector.

Proof of validity. By assumption, the three points A,B,C do not lie on a line. Byconstruction, points A and D lie on different sides of line BC. Hence the segment ADintersects line BC, say at point M . In step (1) and (2), we get three congruent triangles.

Step (1): We confirm that �AMB ∼= �DMB. The matching pieces used for the proofare stressed.

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Figure 5.53: The first pair of congruent triangles

Answer. Indeed, by construction, ∠ABC ∼= ∠DBC. Hence ∠ABM ∼= ∠DBM . (It

does not matter whether M lies on the ray−−→BC or the opposite ray.) Too, we have a

pair of congruent adjacent sides: Indeed BD ∼= BA by construction, and BM ∼= BM .Now SAS congruence implies �AMB ∼= �DMB.

Question. Explain carefully why ∠AMB is a right angle.

Answer. Because of �AMB ∼= �DMB, we get ∠AMB ∼= ∠DMB. Because point Mlies between A and D, these are two supplementary angles. Hence they are right angles.

Step (2): Finally, we confirm that �AMB ∼= �AMC. Again the pieces needed tomatch for this theorem are stressed in the drawing.

Figure 5.54: The second pair of congruent triangles

Answer. Indeed, I use the hypothenuse-leg theorem. ∠AMB ∼= ∠AMC ∼= R, becausea right angle is congruent to its supplement. (Again, it does not matter whether M lies

on the ray−−→BC or the opposite ray.) Too, we have a pair of congruent sides: Indeed

AB ∼= AC by construction, and AM ∼= AM .

From the triangle congruence �AMB ∼= �AMC, we get ∠MAB ∼= ∠MAC, andMB ∼= MC. Since point M lies on the line BC, the last congruence shows that M lies

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between B and C, too. Hence ray−−→AM =

−−→AD lies inside the given ∠BAC and bisects

this angle.

Proposition 5.47 (Existence of the angular bisector). For any angle ∠BAC, thereexists a ray

−−→AD inside the given angle such that ∠DAB ∼= ∠DAC.

Proposition 5.48 (The Hinge Theorem). [Euclid I.24] Increasing the angle betweentwo constant sides increases the opposite side of a triangle.

Figure 5.55: The Hinge Theorem

Corollary 10 states the equivalence we get by taking from [Euclid I.24] and [Euclid I.25]together:

Corollary 10. Given are two triangles with two pairs of congruent sides. In the firsttriangle, the angle between them is smaller, congruent, or greater than in the second oneif and only if the opposite side is smaller, congruent, or greater in the first triangle.

Proof. Given are �ABC and �A′B′C ′ with a = a′ and c = c′. Assuming β < β′, wehave to check whether b < b′. One can assume that A = A′, B = B′, and put the twopoints C and C ′ lie on the same side of line AB. By the hypothesis β < β′, the twopoints A and C ′ lie on different sides of line BC. Hence this line intersects the segmentAC ′.

Indeed, as stated by the Crossbar theorem, the ray−−→BC intersects the segment AC ′.

Let D be the intersection point. We now distinguish three cases:

(a) B ∗D ∗ C. This case occurs if the triangle side BC to be rotated is long enough.

(b) C = D.

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Figure 5.56: The Hinge Theorem: turning a long side

Figure 5.57: The Hinge Theorem: the borderline case

(c) B ∗ C ∗D. In this case the triangle side BC to be rotated is rather short.

In the border line case (b), we see directly that β < β′ implies A ∗ C ∗ C ′ and henceAC < AC ′, as to be shown. In the two other cases, we apply Euclid I.19 to triangle�ACC ′. Thus it is enough to show that this triangle has a larger angle ε = ∠ACC ′ atvertex C than the angle ε′ = ∠AC ′C at vertex C ′. The �BCC ′ is isosceles. Hence, byEuclid I.5, it has two congruent base angles, which we denote by ϕ.

In case (a), we proceed as follows: Because B and C lie on opposite sides of AC ′,comparison of angles at vertex C ′ yields

(1) ε′ < ϕ

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Figure 5.58: The Hinge Theorem: turning a short side

And because A and C ′ lie on opposite sides of line BC, comparison of angles at vertexC yields

(2) ϕ < ε

Because of transitivity, (1)(2) together imply ε′ < ε and hence AC ′ > AC.

In case (c), we show that ε is an obtuse, and ε′ is an acute angle. Because A and C ′ lieon opposite sides of line BCD, we get

(3) ∠C ′CD < ∠C ′CA = ε

but ∠C ′CD is an exterior angle of the isosceles �BCC ′. Because the base angles ofan isosceles triangle are acute, its supplement ∠C ′CD is obtuse. Hence by (3), ε isobtuse, too. Since the triangle �ACC ′ can have at most one right or obtuse angle, theangle ε′ is acute. Now ε′ < R < ε implies again ε′ < ε, and hence Euclid I.19 yieldsAC ′ > AC.

Proposition 5.49 (The Midpoint of a Segment). [Theorem 26 in Hilbert] Anysegment has a midpoint.

Construction 5.7. Given is a segment AB. Transfer congruent angles with the end-points of the given segment AB as vertices, on different sides of AB. Next we transfercongruent segments AC ∼= BD onto the newly produced legs of these two angles. Finally,

lines←→AB and

←→CD intersect at the midpoint M .

(a) Here is a drawing.

(b) Explain why lines AB and CD intersect.

Answer. By construction, points C and D lie on different sides of line AB. Hence,by the plane separation theorem, line AB and segment CD intersect.

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Figure 5.59: Hilbert’s construction of the midpoint

LetM be the intersection point. From the plane separation theorem, too, it followsthat the intersection point M lies between C and D. But it turns out to be harderto see why M lies between A and B!

(c) Show that M = A is impossible.

Figure 5.60: M = A is impossible

Answer. In that case line l = AC would go through point D. The ray−→AC would

be an extension of side AD of the triangle �ABD. This triangle would have theinterior angle ∠ABD congruent to the exterior angle ∠CAB, contradicting theexterior angle theorem. (Remember: an exterior angle is always greater than anonadjacent interior angle.)

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(d) Show that M ∗ A ∗B is impossible.

Figure 5.61: Point A lying between M and B is impossible

Answer 1. We use Pasch’s axiom for �MBD and line l = CA. Which conclusiondo you get?

Answer. The line CA enters �MBD on the side MB. By Pasch’s axiom, thisline either (i) intersects side DB, or (ii) goes through point D, or (iii) intersectsside MD.

In all three cases, we derive a contradiction:

Case (i): suppose line l intersects segment DB, say at point G. The �ABGwould have the interior ∠ABG congruent to the exterior ∠CAB. This con-tradicts the exterior angle theorem, which tells an exterior angle is alwaysgreater than a nonadjacent interior angle.

Case (ii): suppose line l goes through point D. The �ABD would have theinterior ∠ABD congruent to the exterior ∠CAB. This contradicts the ex-terior angle theorem, which tells an exterior angle is always greater than anonadjacent interior angle.

Case (iii): suppose line l intersects segment MD, say at point F . Points C �= Fare different, because they lie on different sides of AB. The lines AC andMD intersect both in point C and in point F ,. Because this are two differentpoints, they determine a line uniquely. Hence all five points C,M,D,A, F lieon one line. Hence we are back to the case M = A, ruled out earlier.

Answer following Hilbert. We apply the exterior angle theorem twice. Here is asketchy drawing for that impossibility. 26 In triangle �ABD, the interior angle

26The drawing, too, occurs in the millenium edition of ”Grundlagen der Geometrie”, page 26.

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Figure 5.62: Point A lying between M and B is impossible

at vertex B is β = ∠DBM , which is smaller than the exterior angle at vertexε = ∠BMC. In triangle �AMC, the angle ε from above is interior angle atvertex M , and hence smaller than the exterior angle α = ∠BAC at vertex A.

Now transitivity yields β < ε < α. On the other hand, the angles ∠DBA = βand α = ∠CAB are congruent by construction. This contradiction rules out thecase M ∗ A ∗B.

(e) Now we know that M lies between A and B, we finally can prove that M is themidpoint.

Question. Which congruence theorem is used for which triangles?

Answer. One uses SAA congruence for�AMC and�BMD. Indeed, the angles at

Figure 5.63: Apply the SAA congruence

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A and B are congruent by construction, and the angles at vertex M are congruentvertical angles. (Both statements are only true because M lies between A and B!)The sides AC and BD opposite to those angles are congruent by construction.Hence the two triangles are congruent, and especially AM ∼= MB

Figure 5.64: The generic situation for Proposition 5.50, for which we prove: Two segmentsCX and DY on different sides of XY are congruent if and only if midpoint M of segment CDlies on line l. The third figure shows the case with both conditions true.

Problem 5.9. Given is a convex quadrilateral with two pairs of congruent oppositesides. Prove that the diagonals bisect each other.

Use the result of problem 5.6 and assume Hilbert’s construction of the midpoint ofa segment is valid. Provide a drawing.

Figure 5.65: The diagonals of a parallelogram bisect each other.

Answer. Since the quadrilateral is convex, the opposite vertices C and D lie on differentsides of the diagonal AB.

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We have proved in problem 5.6 that the angles β = ∠ABC and α = ∠BAD arecongruent. Hence the figure repeats exactly Hilbert’s construction of the midpoint ofsegment AB. Hence the diagonal segments intersect in the midpoint M of diagonal AB.Similarly, we prove that M is the midpoint of the other diagonal CD, too.

Proposition 5.50. The midpoint of a segment lies on a given line l if and only if thetwo endpoints of a segment have the same distance to the line, and lie on the oppositesides of it.

Proof. Assume that the two endpoints C and D of the given segment lie on oppositesides of l. Furthermore, assume CX ∼= DY are the congruent segments to the footpoints X and Y .

Let Q be the intersection point of line l and segment CD, which exists by planeseparation. In the special case X = Y , we are ready immediately. Otherwise, we needto see why Q lies between X and Y ! This is done is the same way as in Hilbert’sconstruction 5.7 of the midpoint, which is now applied to the segment AB = XY .

Finally, one obtains the triangle congruence

(oneflier) �CQX ∼= �DQY

via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,and the congruent segments CX ∼= DY . Hence CQ ∼= DQ, confirming that Q = M isthe midpoint of segment CD.

Conversely, assume that the midpoint M of segment CD lies on the line l. Inthe special case X = Y we are ready immediately. Otherwise, we need to confirm, oncemore, why M lies between X and Y ! Again one needs to use the exterior angle theorem.Finally, one obtains the triangle congruence

(oneflier) �CMX ∼= �DMY

via SAA congruence, using the right angles at X and Y , vertical angles at vertex M ,and the congruent segments CX ∼= DY , as to be shown.

5.8 SSA congruence

Next we study the possibilities and difficulties with SSA congruence. Thus the matchingpieces of the two given triangles are two pairs of sides, and one pair of angles oppositeto one of these sides. I follow Euler’s conventional notation: in triangle �ABC, leta = BC, b = AC, and c = AB be the sides and α := ∠BAC, β := ∠ABC, andγ := ∠ACB be the angles.

Problem 5.10. Investigate an example Use straightedge and compass to constructin Euclidian geometry all triangles (up to congruence) with γ = ∠ACB = 30◦, sideAC = 10 units and side AB given as below. How many non-congruent solutions (none,

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one, two?) do you get in each case? How many of them are acute, right or obtusetriangles? Measure and report the angle β = ∠ABC for all your solutions. Make clearby the drawings what happens in all cases, especially how many acute, right, and obtusetriangles you get as solutions.

Hint: It is convenient to construct all triangles with one common side AC. Checkcarefully to find the obtuse angles! (a) AB = 4 units.

Figure 5.66: A Euclidean example for SSA triangle construction

Answer (a). One has to begin the construction by drawing a segment AC of length 10,

and a ray r with vertex C, forming an angle of 30◦ with←−CA. The point B has to lie on

the ray r, as well as on a circle of radius AB around A. In case (a), the circle does notintersect this ray, hence there is no solution.

(b) AB = 5 units.

Answer. In case (b), the circle just touches the ray r, hence there is one solution, witha right angle at B.

(c) AB = 5.5 units.

Answer. In case (c), the circle intersect the ray r in two points, hence there are two noncongruent solutions. At vertices B and B′, I measure the angles about β = 67◦ andβ′ = 113◦. One solution is an acute, the other an obtuse triangle.

(d) AB = 6 units.

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Answer. Again in case (d), the circle intersects the ray r in two points, and there aretwo non congruent solutions. At vertices B and B′, I measure the angles about β = 57◦

and β′ = 123◦. Both solutions are obtuse triangles. Indeed, the first solution has theobtuse angle α = 180◦ − β − γ = 93◦.

Proposition 5.51 (SSA Matching Proposition). Given are two triangles. Assumethat two sides of the first triangle are pairwise congruent to two sides of the secondtriangle, and that the angles across to one of these pairs are congruent.

Then either the two triangles are congruent, or the angles across the other pair ofcongruent sides add up to two right angles.

Proof. It is assumed that two sides and the angle across one of these sides of �ABCcan be matched to congruent pieces of �A′B′C ′:(SSA) c = AB ∼= A′B′ = c′, b = AC ∼= A′C ′ = b′, γ = ∠ACB ∼= ∠A′C ′B′ = γ′

We have to show that either (a) or (b) holds:

(a) �ABC ∼= �A′B′C ′.(b) The two angles across the second matched side β = ∠ABC and β′ = ∠A′B′C ′ are

(congruent to) supplementary angles. One of them is acute and the other one isobtuse.

We begin by reducing the problem to the special case that A = A′, C = C ′ and thatB and B′ = D lie on the same side of AC. We use the SAS congruence to construct atriangle �ADC, such that �A′B′C ′ ∼= �ADC, with points B and D on the same sideof AC. The drawing below shows that procedure.

Figure 5.67: Matching two triangles with SSA

Question. Explain how you get the triangle �ADC congruent to �A′B′C ′.

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Answer. I just transfer segment C ′B′ onto the the ray−−→CB and get C ′B′ ∼= CD. The

congruence �A′B′C ′ ∼= �ADC, follows from SAS, given in Theorem 12 of Hilbert (seeProposition 5.10 above).

Assume that congruence (a) does not hold. We have to prove that case (b) occurs.We know that α �= α′, since otherwise the SAS-congruence theorem implies �ABC ∼=�A′B′C ′, which again would be case (a) just ruled out. Without loss of generality, wecan assume α′ < α. Since ∠DAC = α′ < α = ∠BAC, the point D lies between B andC, as shown in the drawing.

Since AD ∼= A′B′ ∼= AB by assumption, triangle �ABD is isosceles, with baselineBD. By Euclid I.5, the two base angles of an isosceles triangle are congruent. One ofthem is the angle β = ∠ABD. By Proposition 5.36(ii), these base angles are alwaysacute.

The angle β′ = ∠A′B′C ′ ∼= ∠ADC is the supplement to the second base angle∠ADB ∼= ∠ABD = β. Hence β and β′ are congruent to supplementary angles. Thefirst one is acute, the second one is obtuse, as claimed in (b).

Corollary 11 (SSAA Congruence). Two triangles which have two pairs of congruentsides, and two pairs of congruent angles across to the latter, are congruent.

Proposition 5.52 (Restricted SSA Congruence, case of a unique solution). Asin the matching Proposition above, we assume that two sides and the angle across oneof these sides of �ABC can be matched to congruent pieces of �A′B′C ′:(SSA) AB ∼= A′B′ , AC ∼= A′C ′ and ∠ACB ∼= ∠A′C ′B′

Under the additional assumption that either (i) or (ii) hold,

(i) The given angle γ lies across a side longer or equal to the other given side: c =AB ≥ AC = b.

(ii) The given angle γ = ∠ACB is right or obtuse.

congruence holds among all triangles matched by (SSA). In all these cases there donot exist any non congruent triangles with congruent sides b, c and angle γ. Actuallyassumption (ii) implies (i).

Proof of Proposition 5.52. Assume that (ii) holds. Indeed, by Proposition 5.36(i), atriangle can have at most one angle which is right or obtuse. Hence assumption (ii)implies γ ≥ β. By Euclid I.19, across the greater angle of any triangle lies the longerside. Hence γ ≥ β implies c ≥ b, which is assumption (i).

We now assume that c = AB ≥ AC = b holds, as stated by (i). By Euclid I.18,across the longer side of a triangle lies the greater angle. Hence c ≥ b implies γ ≥ β.A triangle cannot have two angles which are right or obtuse. Hence β and β′ are bothacute or right. Now congruence follows because case (b) in the matching Proposition isruled out.

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Corollary 12 (SSA Congruence for isosceles and right triangles). (i’) Two isosce-les triangles with congruent legs and congruent base angles are congruent.

(ii’) Any two right triangles with congruent hypothenuses and one pair of congruent legsare congruent.

Proof. Assumption (i’) is a special case of (i), and (ii’) a special case of (ii).

Proposition 5.53 (Restricted SSA Congruence Theorem, case with non unique-ness). Again we assume that two sides and the angle across one of these sides of �ABCcan be matched to congruent pieces of �A′B′C ′:(SSA) AB ∼= A′B′ , AC ∼= A′C ′ and ∠ACB ∼= ∠A′C ′B′

Complementary to Proposition 5.52, we assume that neither (i) nor (ii) nor β ∼= R holds.In other words, we assume that the given angle γ = ∠ACB is acute and lies across theshorter given side: c = AB < AC = b, and angle β is not a right angle. Under theseadditional assumptions (a) There exist two non congruent triangles matched by (SSA).

(b) Nevertheless, equivalent are

(1) The two given triangles �ABC and �A′B′C ′ are congruent.(2) (SSAA) The two triangles can be matched in two sides and the two angles across

both given sides.

(3) For both triangles, the angles β and β′′ across the longer given side b are both acuteor both right or both obtuse.

(4) Both triangles are acute, or both are right, or both triangles are obtuse with the twoobtuse angles at corresponding vertices.

In this case just the two given triangles �ABC and �A′B′C ′—not all other triangleswith those given angle and sides γ, c, b—are congruent.

Proof for part (a). We drop the perpendicular from vertex A onto the opposite sideCB. The foot point F �= B, because otherwise β would be a right angle. Transfer

segment FB on the ray opposite to−−→FB to produce segment FB′′ ∼= FB. The two

right triangles �ABF and �AB′′F are congruent by the Hypothenuse-Leg Theorem15. Hence AB ∼= AB′′. Thus we got two non congruent triangles �ABC and �AB′′Cwhich nevertheless satisfy

(SSA”) AB ∼= AB′′ , AC ∼= AC and ∠ACB ∼= ∠ACB′′

For these non congruent triangles, the two angles β and β′′ are supplementary. One ofthem is acute, the other one is obtuse.

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Figure 5.68: Establish two solutions for SSA

Proof for part (b). Clearly (1) implies (2) implies (3) implies (4). If (4) holds, thematching Proposition excludes non congruent solutions.

Indeed, if both triangles are acute, both angles β and β′ are acute. Congruencefollows by the SSA matching Proposition. If both triangles are right, both angles βand β′ are acute or right. In both cases, congruence follows by the SSA matchingProposition.

If both triangles are obtuse, and obtuse angle occurs at corresponding vertices, theneither both angles β and β′ are obtuse, or both angles α and α′ are obtuse. In the secondcase both angles both angles β and β′ are acute. In both cases, congruence follows bythe SSA matching Proposition.

Remark. Earlier on, we have carefully studied triangles with

c = AB = 5 , 5.5 . 6 b = AC = 10 and γ = ∠ACB = 30◦

in Euclidean geometry. In the first case c = 5, one gets a unique right triangle assolution. In the second case c = 5.5 and c = 6, there are two non-congruent solutions,one of which is an acute triangle, the other one is an obtuse triangle with β′ > R.

In the third case c = 6, there are two non-congruent solutions again. But both areobtuse triangles! One gets one triangle with α > R and β < R, as well as a second onewith α′ < R and β′ > R.

5.9 Reflection

Definition 5.14 (Reflection across a Line). Given is a line l. The reflection acrossline l maps an arbitrary point P to a point P ′ meeting the following requirements:

If P lies on the symmetry axis l, we put P ′ := P .If P does not lie on the symmetry axis, the point P ′ is specified by requiring

(i) P and P ′ lie on different sides of l.

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(ii) the lines l and PP ′ are perpendicular, intersecting at F .

(iii) PF ∼= P ′F .

Proposition 5.54. Given are two points P,Q on the same side of l and their imagesP ′, Q′. We show that

(a) PQ ∼= P ′Q′

(b) If the lines m :=←→PQ and l intersect at M , then P ′, Q′ and M lie on a second line

m′.

Question. For this exercise, you can if you like, do drawing and proof, in neutral geom-etry, on your own.

Figure 5.69: Reflection by a line l

Proof of (a). Let F and G be the foot points of the perpendiculars dropped from P andQ onto the symmetry axis l. We use SAS-congruence for the two triangles �QGF and�Q′GF .

Question. Indeed both triangles have a right angle at G—. Explain why.

Answer. ∠QGF is right by definition of a reflection. Since a right angle is congruent toits supplementary angle, the supplementary angle ∠Q′GF is a right angle, too.

—a common side GF and congruent sides GQ ∼= GQ′, by definition of a reflection.Hence SAS congruence implies

(1) �QGF ∼= �Q′GF

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Because of congruence (1), we get

α = ∠QFG ∼= ∠Q′FG ,(2)

QF ∼= Q′F(3)

Since ∠GFP ∼= ∠GFP ′ is a right angle, angle subtraction yields

(4) β = ∠QFP ∼= ∠Q′FP ′

By construction

(5) PF ∼= P ′F

Using (3)(4) and (5), SAS-congruence implies that

(6) �QFP ∼= �Q′FP ′

Hence especially (a) holds.

Proof of (b). We use SAS-congruence for the two triangles �QGM and �Q′GM . In-deed both triangles have a right angle at G: ∠QGM is right by definition of a reflection.Since a right angle is congruent to its supplementary angle, the supplementary angle∠Q′GM is a right angle, too. Furthermore, they have a common side GM and congruentsides GQ ∼= GQ′, by definition of a reflection. Hence SAS congruence implies

�QGM ∼= �Q′GM

For the angle x between the line l of reflection and the given line m, one gets

(8) x = ∠QMG ∼= ∠Q′MG

Now we argue similarly for the two triangles �PFM and �P ′FM . Thus we get

(9) y = ∠PMF ∼= ∠P ′MF = ∠P ′MG

Now by assumption, the three points P,Q and M lie on a line m.

(10) x = ∠QMG = ∠PMF = y

From (8)(9),(10) and transitivity of angle congruence, we get

(11) ∠Q′MG = ∠P ′MG

Now by Hilbert’s axiom, the angle transfer produces a unique ray. Hence−−→MQ′ =

−−→MP ′.

Thus the uniqueness of angle transfer implies that the three points M,P ′ and Q′ lie onone line.

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An alternative for the proof of (b) . (6) above implies

(7) γ = ∠QPF ∼= ∠Q′P ′F

We apply the extended ASA congruence to triangle �PFM and segment FP ′. Theangles to be transferred to the endpoints of that segment are a right angle at F and γat P ′. Transferring produces as second sides of these angles part of the line l and the

ray−−→P ′Q′. Hence those rays intersect, say at point M ′. By ASA-congruence

�MFP ∼= �M ′FP ′

Hence FM ∼= FM ′. The points Q and Q′ and hence M and M ′ lie on the same side ofPP ′. Hence uniqueness of segment transfer implies M = M ′ as to be shown.

Theorem 5.1. The mapping of reflection preserves incidence and maps segments andangles to congruent ones. Hence it is an isometry.

5.10 Restriction of SAS congruence

The SAS axiom restricted to triangles of the same orientation is discussed by Hilbert inan appendix to the Foundations of Geometry.

Note that the triangles �ABC and �A′B′C ′ have the same orientation if either

both point C lies in the left half-plane of ray−→AB and point C ′ lies in the left half-plane

of ray−−→A′B′; or both point C lies in the right half-plane of ray

−→AB and point C ′ lies in the

right half-plane of ray−−→A′B′. Hilbert postulates SAS congruence restricted to triangles

with the same orientation. and two additional axioms:

III.5* Assume that the two triangles �ABC and �A′B′C ′ have the same orientationand the congruences

AB ∼= A′B′ , AC ∼= A′C ′ , ∠BAC ∼= ∠B′A′C ′

hold, then the congruence ∠ABC ∼= ∠A′B′C ′ is also satisfied.

III.6 If the angles ∠(h′, k′) and ∠(h′′, k′′) are both congruent to angle ∠(h, k), then theangles ∠(h′, k′) and ∠(h′′, k′′) are congruent to each other.

III.7 If the rays c and d have the same vertex as the angle ∠(h, k) and lie in the interiorof that angle, then the angles ∠(h, k) and ∠(c, d) are not congruent.

Hilbert then announces the following result without a proof:

Proposition 5.55 (The importance of Euclid I.5). We assume only the restrictedSAS axiom (III.5*), axioms (III.6) and (III.7), and finally we assume Euclid (I.5) to bevalid: the base angles of any isosceles triangle are congruent. These assumptions implythe unrestricted SAS axiom (III.5),—for any triangles independent of their orientation.

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Problem 5.11 (A hard problem, only for fans). Provide some drawings and provethis proposition.

Definition 5.15 (Skew-Hilbert plane). A skew-Hilbert plane is any model for two-dimensional geometry where Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order(II.1) through (II.4), and congruence (III.1) through (III.4) , the restricted SAS-axiom(III.5*), and the axioms (III.6) and (III.7) hold.

Lemma 5.7 (Transfer of a triangle with the same orientation). Any given triangle�ABC can be transferred into either the left or the right half-plane of any given ray r,to obtain a congruent triangle with one vertex at the vertex of this ray, one side lyingon the given ray, and having the same orientation as the original triangle.

Lemma 5.8 (Preparations in a skew-Hilbert plane). The following Propositions arevalid in any skew-Hilbert plane

• The angle congruence is an equivalence relation. Too, Proposition 5.24 holds:angle comparison holds for congruence classes.

• Proposition 5.10 about SAS congruence holds for two triangles with the same ori-entation.

• As stated in Lemma 5.7, the transfer of a triangle is still possible under the re-striction that the orientation is preserved.

• Proposition 5.18 about angle-addition and subtraction (Theorem 15 in Hilbert)remains valid.

• The Exterior Angle Theorem (Euclid (I.16). Theorem 22 in Hilbert, see Proposi-tion 5.34) is valid.

• The SAA-Congruence Theorem, Proposition 5.45 (Theorem 25 in Hilbert) is validfor triangles of the same orientation.

• The midpoint of a segment can be obtained by the method used in Theorem 26 ofHilbert (see Proposition 5.49). Any segment has a midpoint.

Lemma 5.9. The following Propositions are valid in a skew-Hilbert plane where addi-tionally Euclid (I.5) is assumed.

• From the Theorem 5.19 about the symmetric kite, part (i) holds. 27

• Euclid(I.18) (Theorem 23 of Hilbert, see Proposition 5.39 above) is valid: in anytriangle, across the longer side lies the greater angle.

27Part (ii) cannot be recovered at this point but only later.

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• Euclid (I.19) (Proposition 5.40 above) is valid: in any triangle, across the greaterangle lies the longer side.

• We get Euclid (I.6), the converse of Euclid (I.5).

Figure 5.70: To get a rhombus without use of reflection.

Lemma 5.10. The diagonals of the rhombus bisect each other, they bisect the anglesat the vertices of the rhombus, and they are perpendicularly to each other. For anyisosceles triangle �ABC, we can construct a rhombus �ABCD which inherits threeof its vertices from the triangle.

Proof of the Lemma. The midpoint M of side AC can be obtained with Hilbert’s con-struction. From SAS axiom (III.5*) we derive the triangle congruences

�AMB ∼= �CMB′′ and �CMB ∼= �AMB′′

Hence we have obtained a rhombus �ABCB′′. Too, get four congruent angles at verticesA and C and see that the diagonal AC bisects the angles at these two vertices. Beginningwith the isosceles triangle �BAB′′ we repeat a similar argument. Hence the diagonalsof a rhombus bisect each other, and they bisect the angles at the vertices of the rhombus.

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It remains to check that the diagonals are perpendicular to each other. To thisend, we transfer the hypothenuse BC of triangle �BMC on the segment BA andobtain the congruent triangle�BM ′A ∼= �MBC with the same orientation. One needsonly axioms (III.1),(III.4) and the restricted SAS axiom (III.5*). From the first item(i) of the Theorem 5.19 about the symmetric kite, we conclude ∠BMA ∼= ∠BM ′A.From the triangle congruence we conclude we conclude ∠BMC ∼= ∠BM ′A. Hence∠BMC ∼= ∠BMA are congruent supplementary, and hence right angles.

Lemma 5.11. All points P of the perpendicular bisector of two points B and C havecongruent distances BP ∼= CP from these two points.

Proof. We may suppose that point P does not lie on the line BC. We take the smallerone of the two angles ∠CBP and ∠BCP and transfer it to the vertex and onto the sideof the other one. Suppose to be definite that ∠BCP ≤ ∠CBP . We transfer ∠BCP onto

the ray−−→BC. By the Crossbar Theorem, the newly produced ray intersects the segment

BP , say at point Q. The triangle �CQB has congruent base angles, and hence fromEuclid (I.6) we conclude BQ ∼= CQ.

It remains to show that Q = P . Let M be the midpoint of segment BC. From theProposition about the rhombus (Lemma 5.10) we know that line QM is the perpen-dicular bisector of segment BC. Both P and Q are the intersection point of the twodifferent lines QM and BP , hence P = Q. Finally we have confirmed BP ∼= CP .

Lemma 5.12. In a skew-Hilbert plane Euclid (I.5) has been assumed. Theorem 5.19about the symmetric kite, parts (i) and (ii) both hold. Indeed, for a kite with congruentsides XZ1 ∼= XZ2 and Y Z1 ∼= Y Z2, and Z1 and Z2 on different sides of XY , thetriangles �XZ1Y ∼= �XZ2Y are congruent.

End of the proof of the Theorem 5.19 about the symmetric kite. Let Z1 and Z2 be twopoints on different sides of line XY , and assume that XZ1 ∼= XZ2 and Y Z1 ∼= Y Z2.

As explained above, from the isosceles triangle �Z1XZ2, we construct a rhombusZ1XZ2X

′. Its diagonals XX ′ and Z1Z2 intersect perpendicularly in the midpoint M ofsegment Z1Z2.

We begin a similar procedure with the isosceles triangle �Z1Y Z2 and obtain asecond rhombus Z1Y

′Z2Y . Its diagonals Y Y ′ and Z1Z2 intersect perpendicularly in themidpoint M of segment Z1Z2.

Hence all four points X,X ′, Y, Y ′ lie on the perpendicular to segment Z1Z2, erectedat its midpoint M ,—which we recognize as the perpendicular bisector of Z1Z2. Sincewe have shown that the diagonals XX ′ and Y Y ′ bisect the angles at X and Y , we getthe remaining angle congruences for triangles �XZ1Y ∼= �XZ2Y .

Lemma 5.13. Assume that the triangles �APB and �APC lie on different sides ofAP , have congruent sides AB ∼= AC and congruent angles ∠PAB ∼= ∠CAP betweenthese sides and their common side. Then the two triangles are congruent.

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Proof. From the Proposition about the rhombus (Lemma 5.10) we know that line APis the perpendicular bisector of segment BC. From Lemma 5.11, we know that BP ∼=CP . From the Theorem 5.19 about the symmetric kite we conclude that the triangles�APB ∼= �APC are congruent.

End of the proof of the Proposition about importance of Euclid (I.5). Combine Lemma 5.13with the Lemma 5.7 about the transfer of a triangle. Assume the two triangles �ABCand�A′B′C ′ have different orientations, have corresponding congruent sides AB ∼= A′B′

and AC ∼= A′C ′, and congruent angles ∠BAC ∼= ∠B′A′C ′ between them.

We transfer the angle ∠BAC onto the ray−−→A′C ′, at vertex A′, to the same side of

A′C ′ as B′. On the newly produced ray, we transfer segment AB, starting at vertex A′.Thus we get point B0, such that A′B0 ∼= AB. Because of Lemma ?? we get

�ABC ∼= �A′B0C ′

By Lemma 5.13 we conclude that

�A′B′C ′ ∼= �A′B0C ′

Hence �ABC ∼= �A′B′C ′ as to be shown.

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