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