1. Leaving Certificate - Ordinary level
Theorems
Leaving Certificate - Ordinary level
Developed by Pádraic Kavanagh

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3. Theorem 1: The sum of the degree measures of the angles of a triangle is 1800
Given: Triangle
To Prove: Ð1 + Ð2 + Ð3 = 1800 1 2 3
Construction: Draw line through Ð3 parallel to the base 4 5
Proof: Ð3 + Ð4 + Ð5 = 1800 Straight line
Ð1 = Ð4 and Ð2 = Ð5 Alternate angles
Þ Ð3 + Ð1 + Ð2 = 1800
Ð1 + Ð2 + Ð3 = 1800
Q.E.D.
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4. Theorem 2: The opposite sides of a parallelogram have equal lengths.
Given: Parallelogram abcd
To Prove: |ab| = |cd| and |ad| = |bc|
Construction: Diagonal |ac|
Proof: Ð1 = Ð4 Alternate angles
|ac| = |ac| Given
Ð2 = Ð3 Alternate angles ASA Þ D abc = D acd
Þ |ab| = |cd| and |ad| = |bc|
Q.E.D. c b a d 2 3 1 4
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5. 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.
Given: Diagram as shown with |ab| = |bc|
To Prove: |db| = |be|
Construction: Another transversal through b.
Proof: Ð1 = Ð2 Verticially opposite
|ab| = |bc| Given
Ð3 = Ð4 Alternate angles
ASA Þ D dab = D bec
Þ |db| = |be|
Q.E.D. a b c d e 1 2 3 4
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6. 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.
Given: Diagram as shown with line |xy| parallel to base.
To Prove:
|yc|
|ay| =
|xb|
|ax | x y a b c
Construction: Draw in division lines
m equal parts
n equal parts
Proof: |ax | is divided in m equal parts
Þ |ay| is also
|xb| is divided in n equal parts
Þ |yc| is also Theorem 2
|yc|
|ay| =
|xb|
|ax |
Q.E.D.
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7. 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.
Given: Two Triangles with equal angles
To Prove:
|df |
|ac| =
|de|
|ab|
|ef |
|bc| a c b d f e 1 2 3
Construction: Map D def onto D axy, draw in [xy]
Proof: D def mapped onto D axy
Ð1 = Ð4
Þ [xy] is parallel to [bc] x y 4 5
|ay|
|ac| =
|ax|
|ab| Þ
Theorem 3
|df |
|ac| =
|de|
|ab|
Similarly
|ef |
|bc|
Q.E.D.
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8. a b c a b c a b c a b c Quit Menu
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)
Given: Triangle abc
To Prove: a2 + b2 = c2 a b c a b c
Construction: Three right angled triangles as shown
Proof: Area of large sq. = area of small sq. + 4(area D) (a + b)2 = c2 + 4(½ab)
a2 + 2ab +b2 = c2 + 2ab
a2 + b2 = c2
Q.E.D. a b c a b c
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9. 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)
Given: Triangle abc with |ac|2 = |ab|2 + |bc|2
Construction: Triangle def with |de| = |ab|
|ef | = |bc|
& Ð2 = 900 a b c 1 e d f
To Prove: To prove Ð1 = 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
Þ |ac| = |df|
SSS Þ D abc = D def
Ð1 = 900
Q.E.D. 2
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10. Theorem 8:. The products of the lengths of the sides of a triangle by
Theorem 8: The products of the lengths of the sides of a triangle by the corresponding altitudes are equal.
Given: Triangle abc
To Prove: |bc||ad| = |ac||be|
Construction: Altitudes [be] and [ad]
Proof: Compare D bce to D acd 4 5 1 c d a 3 2 b e a b c 4 5 1 3 2 d e
Ð1 = Ð1
Ð2 = Ð4
Þ Ð3 = Ð5
Similar triangles
|ad|
|be| =
|ac|
|bc|
Þ |bc||ad| = |ac||be|
Q.E.D.
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11. 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.
Given: Triangle abc
To Prove: Ðabc > Ðacb
Construction: Draw [bd] such that |ab| = |ad|
Proof: Ð1 = Ð2 As |ab|= |ad|
Þ Ð1 + Ð3 > Ð2
But Ð2 > Ð4
Þ Ð1 + Ð3 > Ð4
Þ Ðabc > Ðacb
and the greater angle is
opposite the longer side.
Q.E.D a b c d 3 1 2 4
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12. Theorem 10: The sum of the lengths of any two sides of a triangle is greater than that of the third side.
Given: Triangle abc
To Prove: |ba| + |ac| > |bc|
Construction: Draw D acd such that |ad| = |ac|
Proof: In D acd |ad| = |ac|
Ð1 = Ð2
Þ Ð1 + Ð3 > Ð2
Þ |bd| > |bc|
But |bd| = |ba| + |ad|
|bd| = |ba| + |ac| as |ad| = |bc|
Þ |ba| + |ac| > |bc|
Q.E.D. 3 d 1 2 b a c
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