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.

Let PA and QB be the two parallel tangents of the circle with centre O and PQ is the third tangent:

In △OAP and △OCP,

⇒ OA = OC [Radius of same circle]

⇒ OP = OP [Common]

⇒ PA = PC [∵ Tangents from exterior point P are equal in length.]

∴ △OAP ≅ △OCP [By SSS axiom]

∴ ∠APO = ∠CPO [By C.P.C.T.]

⇒ ∠APC = ∠APO + ∠CPO

⇒ ∠APC = 2∠CPO ………..(1)

In △OBQ and △OCQ,

⇒ OB = OC [Radius of same circle]

⇒ OQ = OQ [Common]

⇒ QB = QC [∵ Tangents from exterior point P are equal in length.]

∴ △OBQ ≅ △OCQ [By SSS axiom]

∴ ∠CQO = ∠BQO [By C.P.C.T.]

⇒ ∠CQB = ∠CQO + ∠BQO

⇒ ∠CQB = 2∠CQO ………..(2)

As, sum of co-interior angles between parallel lines = 180°.

⇒ ∠APC + ∠CQB = 180°

⇒ 2∠CPO + 2∠CQO = 180° [From (1) and (2)]

⇒ ∠CPO + ∠CQO = 90° ……….(3)

In △POQ,

⇒ ∠CPO + ∠CQO + ∠POQ = 180° [Angle sum property of triangle]

⇒ 90° + ∠POQ = 180°

⇒ ∠POQ = 180° - 90°

⇒ ∠POQ = 90°.

Hence, proved that PQ subtends a right angle at the centre.