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
∴ ∠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.∴ ∠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
7