Question

在△ABC中,AD⊥BC于D,G是AC上任意一点,GE⊥BC于E,EG的延长线与BA的延长线交于F,∠BAD＝∠CAD,

Answer 1

证明：∵AD⊥BC,GE⊥BC,

∴AD//EF,

∴∠CAD=∠AGF,∠BAD=∠F,

∵∠BAD＝∠CAD,

∴∠CAD=∠F,

∴∠AGF＝∠F.