The document proves that in a circle with lines OA, OB, and OP subtending from the center, angle AOB is twice angle APB. Angle OBP is shown to equal the sum of angles OPA and APB. The exterior angle theorem is then applied to show angle ACB is the sum of angles APB and OBP, and angle AOC is equal to angle ACB minus angle OPA, demonstrating angle AOB is twice angle APB.

1. . We have a circle with the centre O. O Lines OA and OB subtend from the centre O. BP and AP subtend from the circumference. Now, let’s draw some lines. We want to prove that ∠AOB is twice that of ∠APB. P C A B

2. . Lets draw another line OP and let ∠OPA be α˚. Let ∠APB be β˚. O P α C β A B

3. . O Lines OA, OP and OB are equivalent in length as they are all the radii of the circle. P α C β A B

4. . O Therefore, ∠OBP = α+β Also, ∠OAP = α P α C β α A B α+β

5. . O ∠ACB = β+(α+β) = 2β+α Because it is an exterior angle of the triangle CBP! P α C β α 2β+α A B α+β

6. . O ∠AOC = (2β+α)- α = 2β Because ∠ACB is an exterior angle of triangle OCD. And now, the property is proven! Since ∠APB = β, and ∠AOB = 2β, therefore ∠AOB = 2 x ∠APB. P α C β 2β α 2β+α A B α+β

7. A SUMMARY . ∠OBP = α+β (base angles of isosceles triangle) O P ∠ACB = β+(α+β) (exterior angle) = 2β+α α C β 2β ∠OAP = α (base angles of isosceles triangle) α 2β+α ∠AOB = (2β+α)-α (exterior angle) = 2β A B ∠AOB = 2β AND ∠APB = β Therefore, ∠AOB = 2 x ∠AOB α+β http://wingnutmodels.com/images/chalkboard.jpg