In the given figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY and X'Y' at B. Prove that ∠AOB = 90°

In ΔOPA and ΔOCA
∠P ≅ ∠C (Tangent is ⊥ to radius)
seg PA ≅ seg AC (Tangent from external point)
seg OA ≅ seg OA (common side)
∴ ΔOPA ≅ ΔOCA (RHS test of congruence)
∠POA = ∠COA = x (c.a.c.t) (1)
Similarly, ΔOCB ≅ ΔOQB
∠COB = ∠QOB = y (c.a.c.t) (2)
Since P-O-Q, ∠POQ = 180°
∠POA + ∠COA + ∠COB + ∠QOB = 180° (Angle addition property)
∴ x + x + y + y = 180°
∴ 2x + 2y = 180°
∴ x + y = 90°
∠AOB = 90°