Question

In the given figure, chords BA and DC of a circle meet at point P. Prove that:

1. ∠PAD = ∠PCB 
2. PA x PB = PC × D

Answer 1

(i) Prove ∠PAD = ∠PCB

Both ∠PAD and ∠PCB are angles formed by two secants drawn from point P outside the circle.

Each angle intercepts the same arcs AD and CB of the circle.

Angles formed outside a circle by two secants are equal if they intercept the same arcs.

Therefore,

∠PAD = ∠PCB.

(ii) Prove PA × PB = PC × PD

Secant PAB

Secant PCD

By the Secant–Secant Theorem (Power of a Point):

(external segment) × (whole secant) is equal for both.

Therefore, PA × PB = PC × PD.