Question

In the given figure PQ and RS are two parallel tangents to a circle with Centre O and another tangent xy with point of contact C interest PQ at A AND RS at B
Prove that angleAOB= 90

Answer 1

Answer:

Explanation:

See my friends.... It's so easy.

Hope it helps you..

Answer 2

Answer:

Given:

Figure

PQ||RS

Tangents PQ, RS and XY

To Prove: ∠AOB = 90°

Proof:

As 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° ----------- Eq 1

In ΔBOA,

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

From Eq 1,

∠AOB + 90° = 180°

∠AOB = 90°

Hence Proved.