∵四邊形ABCD是正方形,
∴AB∥DH,
∴∠H=∠BAH,∠B=∠BCH,
∴△BEA∽△CEH,
∴ABCH=AEEH=BEEC=2,
設EC=m,則AB=BC=CD=3m,BE=2m,CH=1.5m,
同樣:△法新社∽△DPH
∴fp:pd=ap:ph=af:dh=1.5m:4.5m=1:3,
設AP=n,PH=3n,AH=4n,AE: eh = 2: 1,EH=43n,
∴PE=53n,
∴AP:PE=3:5,
∴appe=35,fpdp=13;
(2)證明:如圖所示,將AE交流DC的延長線延長到H,
∵四邊形ABCD是正方形,
∴AB∥DH,
∴∠H=∠BAH,∠B=∠BCH,
∴△BEA∽△CEH,
∴ABCH=AEEH=BEEC=2,
設EC=2a,BE=4a,那麽AB=BC=CD=6a,CH=3a,AF=2a,
同樣:△AFP∽△HDP,AFDH = APPH = 29
設AP=2k,PH=9k,
∴AH=11k,
∴EH=113k,
∴PE=163k,
∴APPE=38,
∴8ap=3pe;
(3)當AE⊥DF,Tan ∠ BAE = PF: AP = BE: AB = 2: 3時
∫△AFP∽△AFD,
∴FP:AP=AF:AD=2:3,
∴AF=23AD=23AB,BF=13AB,
∴BF=12AF,
∴n=12.