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

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

Solution

Given: XY and X'Y' are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X'Y' at B. To prove: ∠AOB = 90º . Construction: Join OC. Proof: In ΔOPA and ΔOCA OP = OC (Radii ) AP = AC (Tangents from point A) AO = AO (Common ) ΔOPA ≅ ΔOCA (By SSS criterion)

$\therefore$ ∠POA = ∠COA .... (1) (By C.P.C.T)

Similarly, ΔOQB ≅ ΔOCB ∠QOB = ∠COB .......(2) POQ is a diameter of the circle. Hence, it is a straight line.

$\therefore$ ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (1) and (2), it can be observed that 2∠COA + 2∠COB = 180° ∴ ∠COA + ∠COB = 90°

∴ ∠AOB = 90°.

Mathematics

RD Sharma

Standard X