CF is perpendicular to AP

∴∠AEB=∠AFC=90

∠∠AEB = 90°

∴∠EAB+∠EBA=90

∫∠BAC = 90

∴∠EAB+∠ACF=90

∴∠EBA=∠ACF

At △AEB and △CFA:

∠AEB =∠ AFC

∠EBA=∠ACF

AB=AC

∴△AEB is all equal to△△△ CFA.

∴BE=AF

AE=CF

So EF=AE+AF=CF+BE.

You can do it with congruence. It's simple.

I just graduated from the third grade and hope to adopt.