From: Jade, a secondary school student

From a point P outside a circle with centre O, tangents are drawn to meet the circle at A and B.
a) Prove that PO is the right bisector of the chord AB.
b) Prove that <APB = 2<OAB.


Hi Jade.

I've drawn a diagram and labelled some extra points and lines to help us.

1. PAO is a right angle. This is because line segment AP is tangent to the circle. Tangent lines are always perpendicular to the radius (in this case, the radius is segment AO). Due to symmetry, the same goes for PBO.

2. PAO is congruent to PBO. This is because they are two right triangles and two sides are congruent (AO is congruent to BO because both are radii and PO is congruent to itself of course).

3. APO is congruent to BPO and AP = BP. Corresponding parts of congruent triangles are congruent (CPCTC).

4. APX is congruent to BPX. This is because of Side-Angle-Side (S.A.S.): AP, APX, & PX are congruent to BP, BPX, & PX.

5. AX is congruent to BX. Reason: CPCTC. This means PO is bisecting AB.

6. AXP is congruent to BXP. Reason: CPCTC. And since these are supplementary angles, they add up to 180 degrees. This means both of these angles are right angles. So PO is the right bisector of AB.

7. OAX is the complementary angle of AOX. This is because AOX is a right triangle.

8. APO is the complementary angle of AOP (which is also called AOX). This is because APO is a right triangle.

9. APO is congruent to OAX. Because two angles complementary to a third angle are congruent.

10. APB is twice the size of OAB. This is because APB = APO + BPO (that's how it's composed) and APO is congruent to BPO due to CPCTC.

Hope this helps!
Stephen La Rocque.