Rt⊿ACF≌Rt⊿CBE

Then < f = < e

Extend the EC intersection AF to point G, and get ∠BCE=∠FCG from the same vertex angle.

And ∠ BCE+∠ E = 180-90 = 90, then ∠ FCG+∠ F = 90.

∴∠CGF= 180 -90 =90

Namely EC⊥AF