In Figure, 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°.

Join O and C.
In △OPA and △OCA

1. OP = OC [Radius of the circle]
2. AP = AC [Tangents from point A]
3. AO = AO [Common side]
   △OPA ≅ △COA [SSS congruence criterior]
   ∠POA = ∠COA [By CPCT]
   Similarly, △OQB ≅ △OCB [SSS congruence criterior]
   ∠QOB = ∠OQB [By CPCT]
   POQ is a diameter of the circle.
   ∠POA + ∠COA + ∠QOB = 180°
   2∠COA + 2∠COB = 180°
   → ∠COA + ∠COB = 90°
   ∠AOB = 90°