1. 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.

2. 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.

3. Theorem 6: Each exterior angle of a triangle is equal to the sum of the two interior opposite angles1 2 3 4
To Prove: Ð1 = Ð3 + Ð4
Proof: Ð1 + Ð2 = ………….. Straight line
Ð2 + Ð3 + Ð4 = ………….. Theorem 2.
Þ Ð1 + Ð2 = Ð2 + Ð3 + Ð4
Þ Ð1 = Ð3 + Ð4
Q.E.D.

4. 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

5. 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

6. 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