Prove that the intercept of a tangent between two parallel

Solution:

Given: XY and X′Y′ at are two parallel tangents to the circle with centre O and AB is the tangent at the point C, which intersects XY at A and X′Y′ at B.

To Prove: ∠AOB = 90°

Construction: Let us joint point O to C.

Proof:

In ΔOPA and ΔOCA, we have

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ≅ ΔOCA (SSS congruence criterion)

Therefore, ∠POA = ∠COA ……(i) (C.P.C.T)

Similarly, ΔOQB ≅ ΔOCB ……(ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (i) and (ii), it can be observed that

2∠COA + 2∠COB = 180°

∴ ∠COA + ∠COB = 90°

So, ∠AOB = 90°.