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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 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°
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