In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Question

In Fig. 10.13, XY and X&#x2032;Y&#x2032; are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X&#x2032;Y&#x2032; at B. Prove that &#x2220; AOB = 90&#xB0;.

Philoid · Accepted Answer

Let us join point O to C.

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ≅ ΔOCA (SSS congruence criterion)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (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°

∠AOB = 90°