CMO 2006 Problem 4

In a right angled-triangle $ABC$ , $\angle{ACB} = 90^o$ . Its incircle $O$ meets $BC$ , $AC$ , $AB$ at $D$ , $E$ , $F$ respectively. $AD$ cuts $O$ at $P$ . If $\angle{BPC} = 90^o$ , prove $AE + AP = PD$ .