To Prove: $AP \times PB = CP \times PD$Construction: Join $AD$ and CB.Proof: In $\triangle APD$ and $\triangle CPB$ $\angle A = \angle C$ [angles of the same segment] $\angle C = \angle B$ [angles of the same segment] $\therefore \triangle APD \sim \triangle BPC$ [AA Postulate] $\therefore \frac{AP}{CP} = \frac{PD}{PB}$ [corresponding sides of similar $\Delta s$ are proportional] $\therefore AP \times PB = CP \times PD$