(1) as the auxiliary line FD is easy to know.
BDF=∠BFD=∠DPF=45
∵∠ CPF = 90。
∴∠CPD=∠CPF-∠DPF=45
And ≈CDP =∠DFP.
∴△PFD∽△PDC
∴PF/FD=PD/DC
(2)
According to the conclusion of (1), PF/FD=PD/DC.
According to △APF∽△AFD
= & gtPF/FD=AP/AF
According to the conclusion of (1), PF/FD=PD/DC.
∴pd/dc=ap/af=ap/ae=pd/ec(∵AE = af，DC=EC)
∴ PE/BC can be proved by AP/AE=PD/EC.