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