Question

Two chords BA and CD intersect at point P outside the circle. Prove that ΔPDA ~ ΔPBC.

Answer 1

Given: A circle with two chords BA and CD, which meet at a point P outside the circle.

To prove: ΔPDA ~ ΔPBC

Proof: We know that ABCD is a cyclic quadrilateral.

∵ In a cyclic quadrilateral, an interior angle is equal to the opposite exterior angle.

∴ ∠PDA = ∠PBC   ...(1)

Now, In ΔPDA and ΔPBC

∠APD = ∠BPC   ...(Common angle)

∠PDA = ∠PBC   ...[From equation (1)]

∴ ΔPDA ~ ΔPBC   ...(By AA similarity)

Hence Proved.