#### 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.

#### Solution

Join OP, OQ, OA, OB and OC.

In Δ OAP and Δ OCP

OA = OC (Radii of the same circle)

OP = OP (Common)

PA = PC (Tangents from P)

∴ By side – side – side criterion of congruence,

ΔOAP ≅ Δ OCP (SSS postulate)

The corresponding parts of the congruent triangles are congruent.

⇒ `∠`APO = `∠`CPO (cpct) ………….(i)

Similarly, we can prove that

∴ Δ OCQ ≅ 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°

Is there an error in this question or solution?

Solution 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. Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.