1. (a) Proving Angle Properties of Circles

2. In Geometry, there is a property which states that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference of the same circle. P O However for this proof, it is assumed that we are always able to draw a straight line (i.e. POC) which passes through the centre O, and cuts both ∠AOB and ∠APB. ∠AOB = 2 x ∠APB B A e.g. If ∠AOB = 50°, then according to the property, ∠APB = 25°. C

3. But what you were given this figure where it is impossible to draw a straight line that passes through the centre O and still can cut both ∠AOB and ∠APB? O P B A How do you proof it then?

4. Let us show you how to :D

5. 1 Extend line AO C Draw a line from O to P 2 O P B A

6. ∠AOB + ∠BOP + ∠POC = 180° (adj. angles on a straight line) C O P ∠OBP = ∠OPB (isosceles triangle) B A ∠BOP = 180 ° – 2 x ∠OPB O P

7. ∠AOB + 180 ° – 2 x ∠OPB +∠POC =180 ° C ∠OPB = ∠OPA +∠APB O P O P B A B A ∠AOB – 2(∠OPA + ∠APB) + ∠POC = O ∠AOB – 2∠OPA – 2∠APB + ∠POC = O

8. ∠POC = 2∠OPA [∠OPA = ∠OAP (isosceles triangle), ∠POC = ∠OAP + ∠OPA (external angles)] C C O P O P B A B A ∠AOB - 2 ∠OPA - 2 ∠APB + ∠POC = O

9. Therefore… SOLVING THE EQUATION GIVE US C ∠AOB = 2 ∠APB O P PROVEN!! :D B A

10. The end :D

11. Done by Ming Xuan Cheng Mun Natalie Jerald Xin Yao Xin Yi