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**Leaving Certificate - Ordinary level**

Theorems Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh

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**Click on the theorem you what to revise!**

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**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. Quit Menu

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**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 Quit Menu

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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 Quit Menu

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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. Quit Menu

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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. Quit Menu

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**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 Quit Menu

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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 Quit Menu

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**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. Quit Menu

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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 Quit Menu

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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 Quit Menu

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