Theorems to be proven at JC Higher Level Theorem 4 : Three angles in any triangle add up to 180°. Theorem 6 : Each exterior angle of a triangle is equal to the sum of the two interior opposite angles Theorem 9 : In a parallelogram opposite sides are equal and opposite angle are equal Theorem 14 : Theorem of Pythagoras : In a right angle triangle, the square of the hypotenuse is the sum of the squares of the other two sides Theorem 19: The angle at the centre of the circle standing on a given arc is twice the angle at any point of the circle standing on the same arc.

Theorem 4: Three angles in any triangle add up to 180°C. 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.

Theorem 6: Each exterior angle of a triangle is equal to the sum of the two interior opposite angles 1 2 3 4 To Prove: Ð1 = Ð3 + Ð4 Proof: Ð1 + Ð2 = 1800 ………….. Straight line Ð2 + Ð3 + Ð4 = 1800 ………….. Theorem 2. Þ Ð1 + Ð2 = Ð2 + Ð3 + Ð4 Þ Ð1 = Ð3 + Ð4 Q.E.D.

Theorem 9: In a parallelogram opposite sides are equal and opposite angle are equal Use mouse clicks to see proof Given: Parallelogram abcd c b a d 3 To Prove: |ab| = |cd| and |ad| = |bc| and Ðabc = Ðadc 4 Construction: Draw the diagonal |ac| 1 Proof: In the triangle abc and the triangle adc 2 Ð1 = Ð4 …….. Alternate angles Ð2 = Ð3 ……… Alternate angles |ac| = |ac| …… Common Þ The triangle abc is congruent to the triangle adc ……… ASA = ASA. Þ |ab| = |cd| and |ad| = |bc| and Ðabc = Ðadc Q.E.D

Theorem 14:Theorem of Pythagoras : In a right angle triangle, the square of the hypotenuse is the sum of the squares of the other two sides 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

Theorem 19: The angle at the centre of the circle standing on a given arc is twice the angle at any point of the circle standing on the same arc. a b c o To Prove: | Ðboc | = 2 | Ðbac | r 2 5 Construction: Join a to o and extend to r Proof: In the triangle aob 1 4 3 | oa| = | ob | …… Radii Þ | Ð2 | = | Ð3 | …… Theorem 4 | Ð1 | = | Ð2 | + | Ð3 | …… Theorem 3 Þ | Ð1 | = | Ð2 | + | Ð2 | Þ | Ð1 | = 2| Ð2 | Similarly | Ð4 | = 2| Ð5 | Þ | Ðboc | = 2 | Ðbac | Q.E.D