Chapter 3 Foundations of Geometry 2 Wang Chapter 3. Foundations of Geometry 2 31 Remark: To prove...

Chapter 3

Foundations of Geometry 2

3.1 Triangles, Congruence Relations, SAS Hypothesis

Definition 3.1 A triangle is the union of three segments ( called its side), whose endpoints (called its vertices) are taken, in pairs, from a set of three noncollinear points. Thus,if the vertices of a triangle are A, B, and C, then its sides are AB, BC, AC, and the triangleis then the set defined by AB ∪ BC ∪ AC, denoted by ∆ABC. The angles of ∆ABC are∠A ≡ ∠BAC, ∠B ≡ ∠ABC, and ∠C ≡ ∠ACB.

Equality and CongruenceEqual(=): identically the same as. For examples,1) Two points are equal, as in A = B, we mean they coincide.2) AB = CD only if the set of points AB is the exact same set of points denoted by CD.Congruence(∼=): is an equivalence relation. (to be defined later).Congruence for Segments and angles

25

Yi Wang Chapter 3. Foundations of Geometry 2 26

AB ∼= XY iff AB = XY

∠ABC ∼= ∠XY Z iff m∠ABC = m∠XY Z

Congruence for trianglesNotation: correspondence between two triangles Given ∆ABC and ∆XY Z, write

ABC ↔ XY Z

to mean the correspondence between vertices, sides and angles in the order written.Note: There are possible six ways for one triangle to correspond to another.

ABC ↔ XY Z ABC ↔ XZY ABC ↔ Y XZ

ABC ↔ Y ZX ABC ↔ ZXY ABC ↔ ZY X

Definition 3.2 (Congruence for triangles) If, under some correspondence between thevertices of two triangles, corresponding sides and corresponding angles are congruent, thetriangles are said to be congruent. Thus we write ∆ABC ∼= ∆XY Z whenever

AB ∼= XY , BC ∼= Y Z, AC ∼= XZ,

∠A ∼= ∠X, ∠B ∼= ∠Y, ∠C ∼= ∠Z

Notation: CPCF: Corresponding parts of congruent figures (are congruent).Properties of Congruence1. reflexive law: ∆ABC ∼= ∆ABC2. Symmetry Law: If ∆ABC ∼= ∆XY Z, then ∆XY Z ∼= ∆ABC 3. Transitive Law: If∆ABC ∼= ∆XY Z, and ∆XY Z ∼= ∆UV W , then ∆ABC ∼= ∆UV W .Remark: Euclid did not use the word congruence. Euclid attributed to congruent trianglesthe property that on triangle could be placed precisely on top of another.


3.2 Taxicab Geometry: Geometry without SAS con-

gruence

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3.3 SAS, ASA, SSS Congruence and Perpendicular Bi-

sectors

Question: Can we require fewer than six sets of congruent pairs to determine two trianglesare congruent? The SAS HypothesisUnder the correspondence ABC ↔ XY Z, let two sides and the included angles of ∆ABCbe congruent, respectively, to the corresponding two sides and the included angle of ∆XY Z.Note: This can not be established within the current set of axioms. (See Section 3.2.

Axiom 3.3 (SAS Postulate) If the SAS Hypothesis holds for two triangles under somecorrespondence between their vertices, then the triangles are congruent.

Theorem 3.4 (ASA Theorem) If, under some correspondence, two angles and the in-cluded side of one triangle are congruent to the corresponding angles and included side ofanother, the triangles are congruent under that correspondence.

Proof: Outline of the proof: (Use SAS postulate).

Given ∠A ∼= ∠X, AB ∼= XY , and ∠B ∼= ∠Y . If AC ∼= XZ, then the two triangles arecongruent by SAS. But if not, then either AC > XZ or AC < XZ.1) Assume AC > XZ, then we can find D on AC such that A−D − C and AD ∼= XZ by


the Segment Construction Theorem. This will leads to a contradiction.2) One can argue similarly for the case AC < XZ. 2

Example 3.5 Given: ∠EBA ∼= ∠CBD, AB ∼= BC, ∠A ∼= ∠C.Prove: EB ∼= DB.

Conclusions Justifications(1) ∆ABE ∼= ∆CBD ASA(2) ∴ EB ∼= DB CPCF

Example 3.6


Given: M is the midpoint of CD and EF .Prove: ∠C = ∠D.

Example 3.7 Prove that a line that bisects an angle also bisects any segment perpendicularto it that joins two points on the sides of that angle.

Isosceles Triangle Theorem

Definition 3.8 Isosceles triangle: a triangle having two sides congruent.Legs, base, base angles, vertex, vertex angle.

Lemma 3.9 In ∆ABC, if AC ∼= BC, then ∠A ∼= ∠B.

Proof: By the SAS postulate: ∆CAB ∼= ∆CBA, therefore, corresponding angles ∠CABand ∠CBA are congruent, that is ∠A ∼= ∠B.(You could also construct an angle bisector from the vertex). 2

Theorem 3.10 A triangle is isosceles iff the base angles are congruent.

Proof: Only need to show the converse. By the ASA theorem, ∆CAB ∼= ∠CBA. ThereforeAC ∼= BC.

2

Example 3.11 Solve the following problem in proof writing.


Given: GJ = KM = 2, JK = HJ = HK = 10, with betweenness relations as evident fromthe figure. Prove: HG = HM .

Symmetry

Lemma 3.12 If M is the midpoint of segment AB and line←−→PM is perpendicular to AB,

then PA = PB.

Lemma 3.13 If PA = PB and M is the midpoint of segment AB, then line PM is per-pendicular to segment AB.

Lemma 3.14 If PA = PB and M is the midpoint of segment AB, then line PM bisects∠APB.

Perpendicular Bisectors, Locus

Definition 3.15 The Perpendicular bisector of a segment AB to be the line that bothbisects AB and is perpendicular to it.

Locus: The locus of a point is the path or set of points that is determined by that pointwhen it satisfies certain given properties.

Theorem 3.16 (Perpendicular Bisector Theorem) The set of all points (The locus ofa point which is ) equidistant from two distinct points A and B is the perpendicular bisectorof the segment AB.


Remark: To prove above theorem, actually one needs to prove two sets are equal. Namely,the Locus (set of points, in this theorem, it is the perpendicular bisector) is equal to the setof points which are equidistant from the two distinct points A and B.Proof: (⇒) Let ` be the perpendicular bisector of AB, and assume P ∈ `. By Lemma 3.12,PA = PB.(⇐): Assume P is equidistant from A and B. Let M be the midpoint of AB, then by Lemma

3.13,←−→PM⊥AB. (Next argue

←−→PM = `). But `⊥AB, there can be only one perpendicular to

AB at M . Hence←−→PM = `, and P ∈ `. 2

Theorem 3.17 If, under some correspondence between their vertices, two triangles have thethree sides of one congruent to the corresponding three sides of the other, then the trianglesare congruent under that correspondence.

Proof: class exercise. Note, there are three different cases, but can be reduced to two cases.2

Example 3.18 Name all congruent pairs of distinct segments and angles in the followingfigure.

Example 3.19 In the following figure, we have a four-sided figure with the congruent sidesand right angles as marked.(a) Prove that AD ∼= CD.(b) If, in addition, BC ∼= CD, prove ∠D is a right angle.


Theorem 3.20 (Existence of perpendicular from an external point) Let the line `and point A not on ` be given. Then there exists a unique line m perpendicular to ` passingthrough A.

Proof: (Existence) By construction. Locate B and C on `. let ∠DBC ∼= ∠ABC, BD ∼= BA.It follows that B and C are both equidistant from A and D. (Why) Hence AD⊥BC by theperpendicular bisector theorem.(Uniqueness): Class discussion. 2

3.4 Exterior Angle Inequality

Definition 3.21 Let ∆ABC be given, and suppose D is a point on←→BC such that the be-

tweenness relation B −C −D holds. Then ∠ACD is called an exterior angle of the giventriangle. The angles at A and B of ∆ABC are called opposite interior angles of ∠ACD.


Remark:1) In Euclidean geometry: The measure of an exterior angle of any triangle equals the sumof the measures of the other two opposite interior angles.2) in absolute geometry: the above relationship is not valid in general.

Definition 3.22 Absolute geometry consists of part of Euclidean geometry that includesall the axioms except all references to parallel lines, or results therefrom. Absolute geometryprovides foundations not only for Euclidean geometry, but also for non-Euclidean geometry.

Exterior angles: example provided by Henri PoincarePoincare Model for Absolute Geometry


1. C is a circle with center O;

2. All points inside C are “points” of this geometry. A point on or outside the circle isnot a “point”;

3. “line”: either an straight line through O, cut off by C, or the arc of a circle that makesright angle with C and the arc lies inside C;

4. Betweeness: If on an arc of a circle point Q is between points P and R, then we defineP −Q−R. And the definitions for segments, rays, and angles follow this betweennessdefinition.

5. for example, the “segment” AB shown in the following figure.


6. Angle measure: the angle between two curves defined as the angle measure of the angleformed by the tangents to the two curves at the point of intersection.

7. The geometry inside C satisfies all the axioms for absolute geometry in plane. Forexample: two points determine a unique line.

8. Consider one of the triangles: (see the following figure)


The angle sum of a triangle in this geometry is always less than 180.

9. SAS Postulate holds for Poincare model.


External angles in Spherical Geometry

1. Take the unit sphere S;

2. “points”: all the points on the surface of the sphere;

3. “lines”: all great circles of that sphere; “equator”: the great circle on S that lies in ahorizontal plane;“meridian”: the great circles passing through the north and south poles;

4. distance: ordinary (Euclidean) arc length;

5. angle measure: use the measure of the angle between the tangents to the sides at thevertex of angle

6. the axioms of absolute geometry work in this geometry, such as two points determinea unique line.

7. triangle: (a spherical triangle) is simply one made up of arcs of great circles, pairwiseconnected at the endpoints.

8. SAS Postulate holds for spherical geometry.

9. the angle sum of a triangle is always greater than 180.

Exterior angle of a triangle in absolute geometry


Theorem 3.23 (exterior angle Inequality) An exterior angle of a triangle has anglemeasure greater than that of either opposite interior angle.

Remark: This theorem is true for absolute geometry. That means it is true without theconcept of parallelism. It’s a weak form of its Euclidean counterpart.

Proof: See above figure. 2

Applications

Corrolary 3.24 1) The sum of the measures of any two angles of a triangle is less than180.2) A triangle can have at most one right or obtuse angle.3) The base angles of an isosceles triangle are acute.

Proof: Proof of the first statement. Use exterior angle inequality. 2

Example 3.25 Consider the triangle shown below, with certain angle measures indicated.


(a) Use the exterior Angle Inequality to show that ∠C has measure less than 131.(b) In this example, is the angle sum of ∆ABC equal to, or less than 180? (Do not useEuclidean geometry. )

Example 3.26 If ∠ECD is an exterior angle of ∆EAC and A−B−C −D holds, use theExterior Angle Inequality to find upper and lower bounds for(a) m∠EBC = x(b) m∠BEC = y

Theorem 3.27 (Saccheri-Legendre Theorem) The angle sum of angle triangle can notexceed 180.


Proof: Outline of the proof:

1) Let ∆ABC be any given triangle. Locate the midpoint M of AC then extend BM toE such that BM = ME. Repeat this construction in ∆BEC. Continue this process atinfinitum.

Lemma 3.28 the angle sums of all the new triangles constructed in the process remain con-stant.

2) Assume the angle sum of ∆ABC greater than 180, i.e., there is a constant t > 0, suchthat

m∠A + m∠ABC + m∠BCA = 180 + t

3) The measures of angles at E,F,G, · · · are decreasing. Hence eventually have measure< t.Observe that

θ1 + θ2 + θ3 + · · ·+ θn < m∠ABC

thus, there exists a n large enough, such that

θn < t

4) Assume when θn < t, the corresponding triangle is ∆BCW . By 2) The angle sum of∆BCW = 180 + t, so it follows that

180 + t = m∠WBC + m∠BCW + m∠W < m∠WBC + m∠BCW + t

i.e.,180 < m∠WBC + m∠BCW,

which is a contradiction. 2

Remark: We haven’t yet proven that the angle sum of a triangle is 180. That needs theParallel Postulate.


3.5 The Inequality Theorems

Theorem 3.29 (Scalene Inequality) If one side of a triangle has greater length thananother side, then the angle opposite the longer side has the greater angle measure, and,conversely, the side opposite an angle having the greater measure is the longer side.

Proof: Outline:1) In ∆ABC it is given that AC > AB. Locate D on AC so that AD = AB, and jointpoints B and D.

m∠ABC > m∠1 = m∠2 > m∠C

2)Conversely, Given m∠B > m∠C. Assume AC < AB, then by (1), m∠B < m∠C, whichis a contraction.

2

Corrolary 3.30 (1) If a triangle has an obtuse or right angle, then the side opposite thatangle has the greatest measure.(2) The hypotenuse of a right triangle has measure greater than that of either leg.

Theorem 3.31 If A,B and C are any three distinct points, then AB + BC ≥ AC, withequality only when the points are collinear, and A−B − C.


Proof: Outline:1) When A,B, and C are collinear. Without loss of generality, assume A is to the left of C.Either A−B − C, B − A− C or A− C −B. Discuss each case.2) Consider the noncollinear case.

Extend CB to D such that BD = BA. In ∆DAC, DC = AB + BC > AC by the Scaleneinequality. 2

Corrolary 3.32 (Median Inequality) Suppose that AM is the median to side BC of∆ABC. Then

AM <1

2(AB + AC)

Proof: Exercise. 2

Example 3.33 Find which of the angle measures x, y, z, r, and s indicated in the followingfigure is the leaset. Prove your answer.


Theorem 3.34 (SAS Inequality Theorem) If in ∆ABC and ∆XY Z we have AB =XY,AC = XZ, but m∠A > m∠X, then BC > Y Z, and conversely if BC > Y Z, thenm∠A > m∠X.

This theorem is also called “Hinge” or “Alligator” Theorem. Proof:

Outline: 1) Construct ray−−→AD such that

−→AB − −−→AD − −→AC and ∠BAD ∼= ∠X, with AD =

XZ = AC.2) Construct the angle bisector of ∠DAC; 3) Converse argument is by contradiction.2


Example 3.35 In the following figure, a circle with diameter QR and center O is shown.

If P varies on the circle on either side of←→QR, and if θ = m∠POQ, define the function

f(θ) = PQ, 0 < θ < 180

Explain why f(θ) is an increasing function. (that is , if θ1 < θ2, then f(θ1) < f(θ2).)Observe points P and P ′ in the figure. We need to prove that PQ, which is f(θ1), is lessthan P ′Q, which is f(θ2).

Solution: Use SAS inequality theorem.

3.6 Additional Congruence Criteria

Theorem 3.36 (AAS congruence criterion) If under some correspondence between theirvertices, two angles and a side opposite in one triangle are congruent to the correspondingtwo angles and side of a second triangle, then the triangles are congruent.

Proof: Outline of proof:show that the third angle of one triangle must be congruent to the corresponding angle inthe other triangle. This is done by assuming the contrary and the Exterior Angle Inequality.

2 Noncongruent triangles satisfying SSA Hypothesis

1) When conditions of SSA are given, it may have one, two or none solutions for the triangle.Why?2) for the case when two triangles satisfying the given SSA condition but not congruent, weobviously have the following lemma.

Theorem 3.37 (SSA Theorem) If, under some correspondence between their vertices,two triangles have two pairs of corresponding sides and a pair of corresponding angles con-gruent, and if the triangles are not congruent under this correspondence, then the remainingpair of angles not included by the congruent sides are supplementary.


Corrolary 3.38 If under some correspondence of their vertices, two acute angled triangleshave tow sides and an angle opposite one of them congruent, respectively, to the correspondingtwo sides and angle of the other, the triangles are congruent.

The Right Triangle Congruence Criteria

Corrolary 3.39 (HL Theorem) If two right triangles have the hypotenuse and leg of onecongruent, respectively, to the hypotenuse and leg of the other, the right triangles are con-gruent.

Proof: class discussion. 2

Corrolary 3.40 (HA Theorem) If two right triangles have the hypotenuse and acute an-gle of one congruent, respectively, to the hypotenuse and acute angle of the other, the trianglesare congruent.

Corrolary 3.41 (LA Theorem) If under some correspondence between their vertices, tworight triangles have a leg and acute angle of one congruent, respectively, to the correspondingleg and acute angle of the other, the triangles are congruent.

Corrolary 3.42 (SsA congruence Criterion) Suppose that in ∆ABC and ∆XY Z, AB ∼=XY , BC ∼= Y Z, ∠A ∼= ∠X, and BC ≥ BA. Then ∆ABC ∼= ∆XY Z.


Example 3.43 Given: PQ⊥PR, QS⊥SR, and PQ ∼= QS. Prove: PR ∼= RS.


Definition 3.44 The distance between two geometric objects (set of points) is the distancebetween the the closest points in the two sets.

Definition 3.45 The distance from any point P to a line ` not passing through P is thedistance from P to the foot of the perpendicular Q from P to line `. A point is equidistantfrom two lines iff the distances from the point to the two lines are equal.

Theorem 3.46 The distance form a point P to any point Q in line ` is least when PQ⊥`.


Example 3.47 Prove that if PA = PB and M is the midpoint of AB, then M is equidistant

from rays−→PQ and

−→PR.

3.7 Quadrilaterals

Notation:1: 3: quadrilateral2. 2: Squares or rectangles3. : Parallelogram

Definition 3.48 If A,B, C and D are any four points lying in a plane such that no three ofthem are collinear, and if the points are so situated that no pair of open segments determinedby each pair of points taken in the order A,B,C and D (AB, BC, etc.) have points incommon, then the set

3ABCD ≡ AB ∪BC ∪ CD ∪DA

is a quadrilateral, with vertices A,B,C,D, sides AB, BC, CD, DA, diagonals AC, BD,and angles ∠DAB, ∠ABC, ∠BCD, ∠CDA.


adjacent(or consecutive) sides or anglesopposite sides or angles

Convex quadrilaterals

Definition 3.49 A quadrilateral with its diagonals intersecting at a point that lies betweenopposite vertices.

Properties of a convex quadrilateral

• The diagonals of a convex quadrilateral intersect at an interior point on each diagonal;

• if 3ABCD is a convex quadrilateral, then D lies in the interior of ∠ABC, and similarlyfor the other vertices;

• If A,B, C, and D are consecutive vertices of a convex quadrilateral, then m∠BAD =m∠BAC + m∠CAD.

Congruence criteria for convex quadrilaterals

Definition 3.50 Two quadrilaterals 3ABCD and 3XY ZW are congruent under the cor-respondence ABCD ↔ XY ZW iff all pairs of corresponding sides and angles under thecorrespondence are congruent (i.e., CPCF). Such congruence will be denoted by

3ABCD ∼= 3XY ZW

.


Theorem 3.51 (SASAS Congruence) Suppose that two convex quadrilaterals 3ABCDand 3XY ZW satisfy the SASAS Hypothesis under the correspondence ABCD ↔ XY ZW .That is, three consecutive sides and the the two angles included by those sides of 3ABCDare congruent, respectively, to the corresponding three consecutive sides and two includedangles of 3XY ZW . Then 3ABCD ∼= 3XY ZW .

Proof: We must prove that the remaining corresponding sides and angles of the two quadri-laterals are congruent. 2 Other congruence theorems for convex quadrilaterals are

ASASA TheoremSASAA TheoremSASSS Theorem

Question: Is ASSSS a valid congruence criterion for convex quadrilaterals?Saccheri, Lambert quadrilaterals

Definition 3.52 A rectangle is a convex quadrilateral having four right angles.

Definition 3.53 Let AB be any line segment, and erect two perpendiculars at the endpointsA and B. Mark off points C and D on these perpendiculars so that C and D lie on the dameside of line AB, and BC = AD. The resulting quadrilateral is a Saccheri Quadrilateral.Side AB is called the base, BC and AD the legs, and side CD the summit. The anglesat C and D are called the summit angles.


Saccheri Quadrilateral in non-Euclidean Geometry

Remark:1) A Saccheri Quadrilateral in the Poincare Model has acute summit angles;2) A Saccheri Quadrilateral on the unit sphere has obtuse summit angles;

Lemma 3.54 lines←→BC and

←→Ad in the Saccheri quadrilateral can not meet.

Proof: Because the uniqueness of perpendiculars from an external point in absolute geom-etry. 2

Lemma 3.55 A Saccheri Quadrilateral is convex.

Why?

Theorem 3.56 The summit angles of a Saccheri Quadrilateral are congruent.

Proof: See the following picture. 3DABC ∼= 3CBAD under the correspondence DABC ↔CBAD by SASAS Theorem.


2

Corrolary 3.57 1) The diagonals of a Saccheri Quadrilateral are congruent.2) The line joining the midpoints of the base and summit of a Saccheri Quadrilateral is theperpendicular bisector of both the base and summit.3) If each of the summit angles of a Saccheri Quadrilateral is a right angle, the quadrilateralis a rectangle, and the summit is congruent to the base.

Proof: class discussion. 1) and 3) are trivial. 2

Definition 3.58 (Lambert quadrilateral) A quadrilateral in absolute geometry havingthree right angles is called a lambert Quadrilateral.

Remark: its existence is guaranteed by the above Corollary.

Three possible hypothesis1) Summit angles of a Saccheri Quadrilateral are obtuse;2) Summit angles of a Saccheri Quadrilateral are right angles;3) Summit angles of a Saccheri Quadrilateral are acute;

Theorem 3.59 The Hypothesis of the Obtuse Angle is not valid in absolute geometry.


Outline of the proof: (By construction):

Locate the midpoints M and N of sides AB and AC of any triangle ABC, and draw line

` =←−→MN . Then drop perpendiculars BB′ and CC ′ from B and C to line `.

(1) Show 3BCC ′B′ (called the Saccheri Quadrilateral associated with ∆ABC) is aSaccheri Quadrilateral. (BB′ = CC ′ and congruent summit angles at B and C. )(2) The angle sum of ∆ABC has twice the value of the measure of x of each summit angle.thus

2x ≤ 180

(3) For any Saccheri Quadrilateral there is an associated triangle. How?

Remark: 1) The Hypothesis of the Acute Angle for Saccheri is also false. But it is impossibleto prove this with only the axioms of absolute geometry.

2) The length of the base B′C ′ equals twice the length of MN .

3) The summit of a Saccheri Quadrilateral has length greater than or equal to that of thebase.

4) The line segment joining the midpoints of two sides of a triangle has length less than orequal to one-half of the third side.

3.8 Circles

Definition 3.60 A circle is the set of points in a plane that lies at a positive, fixed distancer from some fixed point O. The number r is called the radius, and the fixed point O iscalled the center of the circle. A point P is said to be interior to the circle, or an interiorpoint, whenever OP < r; if OP > r, then P is said to be an exterior point.


Other terminologiesSee the following picture:

Elementary properties of a circle

• The center of a circle is the midpoint of any diameter.

• The perpendicular bisector of any chord of a circle passes through the center.

• A line passing through the center of a circle and perpendicular to a chord bisects thechord.

• A line passing through the center of a circle and perpendicular to a chord bisects thechord.

• Two congruent central angles subtend congruent chords, and conversely.

• Two chords equidistant from the center of a circle have equal lengths, and conversely.

Circular Arc Measure

Definition 3.61 A minor arc is the intersection of the circle with a central angle and itsinterior, a semicircle is the intersection of the circle with a closed half-plane whose edgepasses through the center of the circle, and a major arc of a circle is the intersection of

the circle and a central angle and its exterior. We define the measure múACB of the arc asfollows: (see the following figure)


Minor Arc Semicircle Major Arc

múACB = m∠AOB múACB = 180 múACB = 360−m∠AOB

Theorem 3.62 (Additivity of Arc Measure) Suppose arcs A1 = úAPB and A2 = úBQCare any two arcs of circle O having just one point B in common and such that their union

A1 ∪A2 = úABC is also an arc. Then m(A1 ∪A2) = mA1 + mA2.

Remark: Observe that if we are given two arcs on a circle, one of them has to be a minorarc.Outline of the proof:We distinguish three cases:

1) when úABC is a minor arc or a semicircle.

2) When úABC is a major arc and úBQC is either a minor arc or a semicircle.

3) When both úABC and úBQC are major arcs. See the following picture:

Example 3.63 Arc úSPT shown in the following picture is a major arc and is the union of

arc úSV R (a minor arc) and arc úRPT (a major arc). Using the angle measures shown in


the figure, determine the measures of each of the three arcs, and verify the additivity in thiscase.

Definition 3.64 A line that meets a circle in two distinct points is a secant of that circle.A line that meets a circle at only one point is called a tangent to the circle, and the pointin common between them is the point of contact, or point of tangency.

Theorem 3.65 (Tangent Theorem) A line is tangent to a circle iff it is perpendicularto the radius at the point of contact.

Proof: 1) Assume a line ` is tangent to a circle with center C at A, then A is the point ofcontact. Need to prove AC⊥`. (How?)2) Conversely, assume AC⊥`, show A is the only contacting point. 2

Corrolary 3.66 If two tangents PA and PB to a circle O from a common external point

P have A and B as the points of contact with the circle, then PA ∼= PB and−→PO bisects

∠APB.

Theorem 3.67 (Secant Theorem) If a line ` passes through an interior point A of acircle, it is a secant of the circle and intersects that circle in precisely two points.

Outline of the proof:1. Use Intermediate Value Theorem to prove the line ` intersects the circle at one point.2. Then use an elementary geometry construction to prove there is another intersectionpoint. (construct congruent triangles).3. Show no other intersection points.

Remark:: 1. The Secant Theorem proves, that a line segment joining a point inside a circlewith a point outside must intersect the circle.


2. In general, this is true for any simple closed curves (called Jordan curve) . (JordanClosed Curve Theorem).

Chapter 3 Foundations of Geometry 2 Wang Chapter 3. Foundations of Geometry 2 31 Remark: To prove...

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