Question

In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that ∠POQ = 90°.

Answer 1

In the given figure, join OR.

In ΔOBP and ΔORP,

PB = PR       ...(Tangents drawn from outside point P)

OB = OR       ...(Radii of the circle)

OP = OP       ...(Common side)

∴ ΔOBP ≅ ΔORP       ...(By SSS)

Therefore, ∠BOP = ∠POR = x       ...(cpct)

In ΔOAQ and ΔORQ,

QA = QR       ...(Tangents drawn from outside point Q)

OA = OR      ...(Radii of the circle)

OQ = OQ       ...(common side)

∴ ΔOAQ  ≅  ΔORQ       ...(By SSS)

Therefore, ∠AOQ = ∠ROQ = y       ...(cpct)

As, AOB is a straight line, given

So, ∠AOB = 180°

∴ x + x + y + y = 180°

⇒ 2(x + y ) = 180°

⇒ x + y = 90°

⇒ ∠POQ = 90°

Hence Proved