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Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at center.

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#### Solution

Consider circle with center ‘O’ and has two parallel tangents through A & B at ends of

diameter.

Let tangents through M intersects the tangents parallel at P and Q required to prove is that ∠POQ = 90°.

From fig. it is clear that ABQP is a quadrilateral

∠A + ∠B = 90° + 90° = 180° [At point of contact tangent & radius are perpendicular]

∠A + ∠B + ∠P + ∠Q = 360° [Angle sum property]

∠P + ∠Q = 360°−180° = 180° …..(i)

At P & Q ∠APO = ∠OPQ =1/2∠𝑃

∠BQO = ∠PQO =`1/2`∠𝑄 in (i)

2∠OPQ + 2 ∠PQO = 180°

∠OPQ + ∠PQO = 90° …. (ii)

In ΔOPQ, ∠OPQ + ∠PQO + ∠POQ = 180° [Angle sum property]

90° + ∠POQ = 180° [from (ii)]

∠POQ = 180° − 90° = 90°

∴ ∠POQ = 90°

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