AE=AF reappears.

therefore

∠BAC=2∠E

∠E= 1/2∠BAC

AB=AC

∠C=∠B

=( 180 -∠BAC)/2

therefore

∠E+∠C = 1/2∠BAC+( 180-∠BAC)/2 = 90

therefore

EF⊥BC