Question

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

Answer 1

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) 

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°.

Answer 2

hope helped
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