Question

In figure PQ and RS are two parallel tangents to a circle with Centre and another
tangent AB with the point of contact C intersecting PO at A and RS a B. Prove that
AOB is equal to 90°

Answer 1

Steps are given above ^

Hope this helps!!

Answer 2

Given:-

• PQ and RS are two parallel tangents to a circle with Centre ,PQ||RS.
• tangent AB with the point of contact C intersecting PO at A and RS a B.

To Prove:-

• ∠AOB = 90°

Proof:-

• AM = AC (Tangents from the same point),
• And OM = OC (Radii)

AMOC is a kite.

Similarly,

BNOC is a kite

• → ∠MAC + ∠NBC = 180° (Co-Interior Angles)
• → ∠MXO + ∠OAC + ∠NBO + ∠OBC = 180°

In a kite, vertex angles are bisected,

• → 2∠OAC + 2∠OBC = 180°
• → ∠OAC + ∠OBC = 90° -----------(1)

In ΔBOA,

• → ∠AOB + ∠AOC + ∠OBC = 180° (Angle Sum Property)

From Equation (1)

• → ∠AOB + 90° = 180°
• → ∠AOB = 90°

!!Hence Proved.!!