(1)证明:连接AC,∵弧AC=弧CE,∴∠CEA=∠CAE.∵∠CEA=∠CBA,∴∠CBA=∠CAE,则 ∠CDF=∠CFD(等量代换,等弧圆周角相等则圆心角相等)∴CD=CF.,
又∵AB是直径, ∴∠ACB=90°,
CP⊥AB,∴∠CAP=∠ACP,
∠CAE=∠ACP,∴AD=CD.∴AD=CF
(2)∵∠ACB=90°,∠CAE=∠ACP,∴∠DCF=∠CFD,∴AD=CD-DF=5/4 ,
∵∠ECB=∠DCP,tan∠ECB=3/4 ,∴tan∠DAP= DP/PA=3/4,
∵ DP^2+PA^2=DA^2,∴DP=3/4 ,PA=1,
∴CP=2,∵∠ACB=90°,CP⊥AB,∴△APC∽△CPB,
∴AP/PC=PC/PB, ∴PB=4.