如图,AB=AC,∠ABC=a,EC=ED,∠CED=2a,P为BD的中点,连AE、PE.(2)当A=60°时求证:AE=2PE(3)当a=多少度时,∠AEP=45°
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![如图,AB=AC,∠ABC=a,EC=ED,∠CED=2a,P为BD的中点,连AE、PE.(2)当A=60°时求证:AE=2PE(3)当a=多少度时,∠AEP=45°](/uploads/image/z/6981793-25-3.jpg?t=%E5%A6%82%E5%9B%BE%2CAB%3DAC%2C%E2%88%A0ABC%3Da%2CEC%3DED%2C%E2%88%A0CED%3D2a%2CP%E4%B8%BABD%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E8%BF%9EAE%E3%80%81PE.%EF%BC%882%EF%BC%89%E5%BD%93A%3D60%C2%B0%E6%97%B6%E6%B1%82%E8%AF%81%3AAE%3D2PE%283%29%E5%BD%93a%3D%E5%A4%9A%E5%B0%91%E5%BA%A6%E6%97%B6%2C%E2%88%A0AEP%3D45%C2%B0)
如图,AB=AC,∠ABC=a,EC=ED,∠CED=2a,P为BD的中点,连AE、PE.(2)当A=60°时求证:AE=2PE(3)当a=多少度时,∠AEP=45°
如图,AB=AC,∠ABC=a,EC=ED,∠CED=2a,P为BD的中点,连AE、PE.(2)当A=60°时求证:AE=2PE(3)当a=多少度时,∠AEP=45°
如图,AB=AC,∠ABC=a,EC=ED,∠CED=2a,P为BD的中点,连AE、PE.(2)当A=60°时求证:AE=2PE(3)当a=多少度时,∠AEP=45°
(2)证明:延长EF到M,使PM=PE,连接BM,AM.
∵PB=DP(已知);PM=PE(所作);∠BPM=∠DPE(对顶角相等)
∴⊿BPM≌⊿DPE(SAS),BM=DE=CE;∠PBM=∠D.
∴∠ABM=360°-∠ABC-∠PBM-∠PBC=300°-∠D-∠PBC;
又∠ACE=540°-∠CAB-∠ABC-∠CED-∠D-∠PBC=300°-∠D-∠PBC.
∴∠ABM=∠ACE;
又AB=AC.故⊿ABM≌⊿ACE(SAS),AM=AE;∠BAM=∠CAE.
∴∠MAE=∠BAC=60°,则⊿MAE为等边三角形.
所以,AE=ME=2PE.
(3)当a=45° 时,∠AEP=45° .
证明:延长EP到M,使PM=PE,连接BM,AM.
同理可证:⊿ABM≌⊿ACE,BM=CE;AM=AE;∠BAM=∠CAE.
∴∠MAE=∠BAC=90°;又AM=AE(已证)
∴∠AEP=∠45°.
A是一个度数 ...............
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