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TheoremsLeaving Certificate - Ordinary level

Developed by Pádraic Kavanagh

42510011 0010 1010 1101 0001 0100 1011

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Theorem 1

Theorem 7

Theorem 6

Theorem 3

Theorem 4

Theorem 5

Theorem 8

Theorem 10

Theorem 9

Theorem 2

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Proof: 3 + 4 + 5 = 1800Straight line

1 = 4 and 2 = 5Alternate angles

3 + 1 + 2 = 1800

1 + 2 + 3 = 1800

Q.E.D.

4 5

Given: Triangle

1 2

3Construction: Draw line through 3 parallel to the base

Theorem 1: The sum of the degree measures of the angles of a triangle is 1800

To Prove: 1 + 2 + 3 = 1800

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2

31

4

Given: Parallelogram abcd

cb

a d

Construction: Diagonal |ac|

Proof: 1 = 4 Alternate angles

|ac| = |ac| Given

2 = 3 Alternate angles

ASA abc = acd

|ab| = |cd| and |ad| = |bc|

Q.E.D.

Theorem 2: The opposite sides of a parallelogram have equal lengths.

To Prove: |ab| = |cd| and |ad| = |bc|

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Given: Diagram as shown with |ab| = |bc|

Proof: 1 = 2 Verticially

opposite

|ab| = |bc| Given

3 = 4 Alternate angles

ASA dab = bec

|db| = |be|

Q.E.D.

a

b

c

d

e

1

2

3

4

Construction: Another transversal through b.

Theorem 3: If three parallel lines make intercepts of equal length on a transversal, then they will also make equal length on any

other transversal.

To Prove: |db| = |be|

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Given: Diagram as shown with line |xy| parallel to base.

x y

a

b c

Proof: |ax | is divided in m equal parts

|ay| is also

|xb| is divided in n equal parts

|yc| is also Theorem 2

m e

qu

al

parts

n e

qual

parts

|yc|

|ay|=

|xb|

|ax |

Construction: Draw in division lines

Q.E.D.

Theorem 4: A line which is parallel to one side-line of a triangle, and cuts a second side, will cut the third side in the same proportion as the second.

To Prove:|yc|

|ay|=

|xb|

|ax |

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Given: Two Triangles with equal angles

Proof: def mapped onto axy

1 = 4

[xy] is parallel to [bc]

Construction: Map def onto axy, draw in [xy]

|df |

|ac|=

|de|

|ab|Similarly

|ef |

|bc|=

Q.E.D.

a

cb

d

fe1

2

3

1 3

2

x y4 5

|ay|

|ac|=

|ax|

|ab| Theorem 3

Theorem 5: If the three angles of one triangle have degree-measures equal, respectively, to the degree-measures of the angles of a second triangle, then the lengths of the corresponding sides of the two triangles are proportional.

To Prove:|df |

|ac|=

|de|

|ab|

|ef |

|bc|=

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Given: Triangle abc

Proof: Area of large sq. = area of small sq. + 4(area )

(a + b)2 = c2 + 4(½ab)

a2 + 2ab +b2 = c2 + 2ab

a2 + b2 = c2

Q.E.D.

a

b

c

a

bc

a

b

c

a

b c

Construction: Three right angled triangles as shown

Theorem 6: In a right-angled triangle, the square of the length of the side opposite to the right-angle is equal to the sum of the

squares of the lengths of the other two sides. (Pythagoras)

To Prove: a2 + b2 = c2

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Given: Triangle abc with |ac|2 = |ab|2 + |bc|2

To Prove: To prove 1 = 900

Proof: |ac|2 = |ab|2 + |bc|2 given

|ac|2 = |de|2 + |ef|2 from

construction |ac|2 = |df|2

|df|2 = |de|2 + |ef|2

|ac| = |df|

SSS abc = def

1 = 900

Q.E.D.

e

d

f

a

b c1 2

Construction:Triangle def with|de| = |ab|

|ef | = |bc|

& 2 = 900

Theorem 7: If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the

triangle has a right-angle and this is opposite the longest side. (Converse of Pythagoras’ Theorem)

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Given: Triangle abc

Construction: Altitudes [be] and [ad]

Proof: Compare bce to acd

1 = 1

2 = 4

3 = 5

Similar triangles

|ad|

|be|=

|ac|

|bc|

|bc||ad| = |ac||be|

Q.E.D.

a

b c4

5

13

2

d

e

Theorem 8: The products of the lengths of the sides of a triangle by the corresponding altitudes are equal.

4

5

1cd

a

13

2

b c

e

To Prove: |bc||ad| = |ac||be|

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Given: Triangle abc

Construction:Draw [bd] such that |ab| = |ad|

a

b c

d

31

2

4

Proof: 1 = 2 As |ab|= |ad|

1 + 3 > 2

But 2 > 4

1 + 3 > 4

abc > acb

and the greater angle is

opposite the longer side.

Q.E.D

Theorem 9: If lengths of two sides of a triangle are unequal, then the degree-measures of the angles opposite to them are unequal,

with the greater angle opposite to the longer side.

To Prove: abc > acb

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Given: Triangle abc

Construction:Draw acd such that |ad| = |ac|

Proof: In acd |ad| = |ac|

1 = 2

1 + 3 > 2

|bd| > |bc|

But |bd| = |ba|+ |ad|

|bd| = |ba| + |ac| as |ad| = |

bc|

|ba| + |ac| > |bc|

Q.E.D.

Theorem 10: The sum of the lengths of any two sides of a triangle is greater than that of the third side.

b

a

c

To Prove: |ba| + |ac| > |bc|

3

d

1

2

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