The angle AFD=2EAB=AED, then F is the intersection of the circles passing through ADE and CD.

Manufacturing (as shown in the figure or attached)

Even AF? Make EG⊥AF in g, even CG, BG, EF.

From ∠EGF+ECF=90+90= 180, we get e, g, f, C*** cycles, so ∠EGC=EFC. Similarly, there is ∠BGE=BAE.

And BAE+EFC = 90? So ∠EGC+BGE=90, which means ∠BGC=90.

And e is the midpoint of BC, so BE=EG=CE, and it is easy to prove that △BAE is equal to △GAE△EGF is equal to △ECF.

So AG=AB=CD, GF=CF so AF=AG+GFCD+CF