Question

Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.

Answer 1

Join OP, OQ, OA, OB and OC.

In ΔOAP and ΔOCP

OA = OC  ...(Radii of the same circle)

OP = OP  ...(Common)

PA = PC   ...(Tangents from P)

∴ By side – side – side criterion of congruence,

ΔOAP ≅ ΔOCP  ...(SSS postulate)

The corresponding parts of the congruent triangles are congruent.

`=>` ∠APO = ∠CPO     (c.p.c.t) ...(i)

Similarly, we can prove that

∴ ΔOCQ ≅ ΔOBQ

`=>` ∠CQO = ∠BQO   ...(ii)

∴ ∠APC = 2∠CPO and ∠CQB = 2∠CQO

But,

∠APC = ∠CQB  = 180°

(Sum of interior angles of a transversal)

∴ 2∠CPO + 2∠CQO = 180°

`=>` ∠CPO + ∠CQO = 90°

Now in ΔPOQ,

∠CPO + ∠CQO + ∠POQ = 180°

`=>` 90° + ∠POQ  = 180°

∴ ∠POQ = 90°