AB=AC,AF=AG,AF⊥BD,交BD延长线于F,AG⊥CE交CE延长线于G求证AD=AE
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![AB=AC,AF=AG,AF⊥BD,交BD延长线于F,AG⊥CE交CE延长线于G求证AD=AE](/uploads/image/z/11550717-45-7.jpg?t=AB%3DAC%2CAF%3DAG%2CAF%E2%8A%A5BD%2C%E4%BA%A4BD%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8EF%2CAG%E2%8A%A5CE%E4%BA%A4CE%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8EG%E6%B1%82%E8%AF%81AD%3DAE)
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AB=AC,AF=AG,AF⊥BD,交BD延长线于F,AG⊥CE交CE延长线于G求证AD=AE
AB=AC,AF=AG,AF⊥BD,交BD延长线于F,AG⊥CE交CE延长线于G求证AD=AE
AB=AC,AF=AG,AF⊥BD,交BD延长线于F,AG⊥CE交CE延长线于G求证AD=AE
证明:
因为:AB=AC,AF=AG,又∠AGC=∠AFB=90度,
则 ∠GAC=∠BAF
所以 ∠GAE=∠FAD
又因为 AG=AF ∠AGC=∠AFB=90度
所以 AD=AE
先证△AGC与△AFB全等
得到∠B=∠C
BF=CA
再证△ABD和△ACE全等
就能得出AD=AE