EC=EH, ∠GEC=∠GEH, then: EG⊥CH. (isosceles triangle "three lines in one")

∴∠hce+∠gec=90；

So: ∠HCE=∠AEB, CH ∨ AE;

Ah ∨ CH again; AC⊥EH. Then the quadrangle AECH is a diamond.

Then: ∠BAE =∠FAE =∠FAH = 30∠BAC = 60