∵∠BAC = 90°, and AP is the center line in △AEP and △CPF.

∴AP=CP∠EPA=∠CPF

∵ isosceles triangle triad AP=CP

∴AP sharing ∠ BAC ∠ EAP = ∠ C.

∴∠EAP=45 ∴△AEP≌△CPF

AB = AC，∠BAC=90？ ∴EP=FP, △PEF is an isosceles triangle.

∴∠ C = 45 (in any case, the center line of the connection can prove △ AEP △ CPF).

∵AP⊥BC

∴∠FPC+∠APE=∠APC=90

≈EPA+≈APE = 90

∴∠FPC=∠EPA