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 intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.

Draw a line between points O and C.

In Δ OPA and Δ OCA,

⇒ OP = OC (Radii of the circle)

⇒ AP = AC (The lengths of tangents drawn from an external point to a circle are always equal.)

⇒ AO = AO (Common)

∴ Δ OPA ≅ Δ OCA (By SSS axiom)

By C.P.C.T.,

∠POA = ∠AOC ……….. (1)

In Δ COB and Δ BOQ,

⇒ OQ = OC (Radii of the circle)

⇒ BC = BQ (The lengths of tangents drawn from an external point to a circle are always equal.)

⇒ OB = OB (Common)

∴ Δ COB ≅ Δ BOQ (By SSS axiom)

By C.P.C.T.

∴ ∠COB = ∠BOQ ……….. (2)

PQ is a diameter, hence a straight line and ∠POQ = 180°

From figure,

⇒ ∠POQ = ∠POA + ∠AOC + ∠COB + ∠BOQ

⇒ ∠POA + ∠AOC + ∠COB + ∠BOQ = 180°

⇒ ∠AOC + ∠AOC + ∠COB + ∠COB = 180° [From equation (1) and (2)]

⇒ 2∠AOC + 2∠COB = 180°

⇒ 2(∠AOC + ∠COB) = 180°

⇒ ∠AOC + ∠COB = 90°

From the figure,

⇒∠AOC + ∠COB = ∠AOB

∴ ∠AOB = 90°

Hence proved that ∠AOB = 90°.