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Classical problems of mathematics and geometry in senior high school entrance examination
Proof: ∠ AEB = ∠ AEF; ∠GEC=∠GEH, then: ∠ AEB+∠ GEC = 90;

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