In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that

(i) ∠PAD = ∠PCB

(ii) PA × PB = PC × PD.

(i) From figure,

∠PAD + ∠DAB = ∠PCB + ∠BCD (∵ both are equal to 180°)….(i)

∠DAB = ∠BCD (∵ angles in same segment are equal.)

Putting this value of ∠DAB in (i) we get,

⇒ ∠PAD + ∠BCD = ∠PCB + ∠BCD
⇒ ∠PAD = ∠PCB + ∠BCD - ∠BCD
⇒ ∠PAD = ∠PCB

Hence, proved that ∠PAD = ∠PCB.

(ii) In △PBC and △PAD,

∠PAD = ∠PCB (Proved above.)

∠P = ∠P (Common angle.)

△PBC ~ △PAD. (By AA axiom)

Since triangles are similar hence, the ratio of corresponding sides are similar,

$\therefore \dfrac{PC}{PA} = \dfrac{PB}{PD} \\[1em] \Rightarrow PA \times PB = PC \times PD.$

Hence proved that PA × PB = PC × PD.