∵CD⊥AB, that is, △BCD is a right triangle.

E is the midpoint of the hypotenuse BC of Rt△BCD.

∴DE= 1/2BC

Use c as CG∨DF and g as AB cross.

∵ is the midpoint of BC

∴DE is the middle line of △BCG.

∴DE= 1/2CG

∴BC=CG

Also ∵CG∨DF

∴△ACG∽△AFD

∴AC/AF=CG/DF=BC/DF

That is AF/DF=AC/BC.

I only know the first question, sorry ~ ~