初二数学题,如图AB=AC,AD=AE,AB⊥AC,AD⊥AE.求证∠B=∠C,BD=CE
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![初二数学题,如图AB=AC,AD=AE,AB⊥AC,AD⊥AE.求证∠B=∠C,BD=CE](/uploads/image/z/6071665-49-5.jpg?t=%E5%88%9D%E4%BA%8C%E6%95%B0%E5%AD%A6%E9%A2%98%2C%E5%A6%82%E5%9B%BEAB%3DAC%2CAD%3DAE%2CAB%E2%8A%A5AC%2CAD%E2%8A%A5AE.%E6%B1%82%E8%AF%81%E2%88%A0B%3D%E2%88%A0C%2CBD%3DCE)
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初二数学题,如图AB=AC,AD=AE,AB⊥AC,AD⊥AE.求证∠B=∠C,BD=CE
初二数学题,如图AB=AC,AD=AE,AB⊥AC,AD⊥AE.求证∠B=∠C,BD=CE
初二数学题,如图AB=AC,AD=AE,AB⊥AC,AD⊥AE.求证∠B=∠C,BD=CE
∵AB⊥AC,AD⊥AE,
∴∠BAC=∠DAE,
∴∠CAD+∠BAC=∠DAE+∠CAD
即∠BAD=∠CAE
在△ABD和△ACE中
AB=AC
∠BAD=∠CAE
AD=AE
∴△ABD≌△ACE(SAS)
∴∠B=∠C,BD=CE (全等三角形中对应边相等)
∠BAD=∠CAD+90度
∠CAE=∠CAD+90度
所以∠BAD=∠CAE
又AB=AC,AD=AE
所以根据边角边法则,三角形ABD和三角形ACE全等
所以∠B=∠C, BD=CE
由题目可得,∵AB=AC ∠BAD=∠CAE AD=AE ∴△ABD≌△ACE(SAS)
由全等三角形定理可得 ∠B=∠C,BD=CE
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