Question

AB is the diameter of a circle with centre O. A line PQ touches the given circle at point R and cuts the tangents to the circle through A and B at point P and Q respectively.
Prove that : angle POQ=90°

Answer 1

Given that AQ, BP, and QP are tangents, we know that

angles QAO, QRO, PBO, and PRO are right angles.

We know that OA, OB, and OR are radii and are therefore

equal.

We can now use Pythagorean theorem to show that AQ = RQ, and BP = RP

Thus, by the SSS postulate, triangle AQO is congruent to triangle AREA, and triangle BPO is congruent to triangle

RPO.

This now gives us angle AOQ = angle ROQ, and angle BOP = angle ROP

Since AOQ + ROQ + BOP + ROP = 180

We have

ROP + ROP + ROP + ROP = 180

2ROQ + 2ROP = 180

ROQ + ROP = 90

POQ = 90

And OP and OQ are perpendicular.