△ABC,分别以AB、AC为边在△ABC外侧作△ABE和△ACD,使AB=AE,AC=AD,且∠BAE=∠CAD,BD与CE交于点H

证明：在EH上截取EF=BH,连AF,因为∠BAE=∠CAD所以∠EAB+∠BAC=∠CAD+∠BAC即∠EAC=∠BAD又AE=AB,AC=AD所以△AEC≌△ABD所以∠AEC=∠ABD因为AE=AB,EF=BH所以△AEF≌△ABH所以AF=AH,∠EAF=∠BAH因为∠EAF+∠FAB=60所以...