∵∠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