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2014-10-25
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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.
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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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.
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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