如图,已知在△ABC中,∠BAC=90°,AB=AC,AD⊥BC于D,E为AC上一点,BE交AD于点H,AF⊥BE于G,
1个回答

1)证明:AB=AC,∠BAC=90°,AD垂直BC,则AD=BC/2=BD;

又AF垂直BE,则∠DAF=∠DBH(均为∠DFG的余角);

又∠ADF=∠BDH=90度.故⊿ADF≌⊿BDH,得DH=DF.

2)DH=DF的结论依然成立.

证明:AB=AC,∠BAC=90°,AD垂直BC,则AD=BC/2=BD;

AF垂直BE,则∠DAF=∠DBH(均为∠DHG的余角);

又∠ADF=∠BDH=90度.故⊿ADF≌⊿BDH,得DH=DF.

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