如图,已知AD,BD分别平分∠EAB和∠CBA,EB过点D,AB=AE+BC.求证:AE平行BC
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如图,已知AD,BD分别平分∠EAB和∠CBA,EB过点D,AB=AE+BC.求证:AE平行BC
先解一个与本题类似的题目,然后再说本题.
1、四边形ABCE中,已知AD,BD分别平分∠EAB和∠CBA,EC过点D,AB=AE+BC.求证:AE平行BC.这里D点是在EC边上.
证明:在AB边上取一点F,并使AF=AE,连接FD,在△AFD与△AED中,
∵AD是公用边,AF=AE,AD平分∠EAF(∠EAB),∴△AFD≌△AED,∠AED=∠AFD……①
∵已知AB=AE+BC,AF=AE,而AB=AF+FB,∴FB=BC,
仿前可证△BFD≌△BCD,得∠DCB=∠DFB……②
①、②两式中∵∠AFD+∠DFB=180°,∴∠AED+∠DCB=180°,于是AE∥BC.
2、回到原题,若D点在EB上,仍在AB上取AF=AE,连接DF、DC,套用上面的证法,得∠AED+∠DCB=180°,从而∠AEC+∠ECB>180°,则AE与BC必不平行.
1、四边形ABCE中,已知AD,BD分别平分∠EAB和∠CBA,EC过点D,AB=AE+BC.求证:AE平行BC.这里D点是在EC边上.
证明:在AB边上取一点F,并使AF=AE,连接FD,在△AFD与△AED中,
∵AD是公用边,AF=AE,AD平分∠EAF(∠EAB),∴△AFD≌△AED,∠AED=∠AFD……①
∵已知AB=AE+BC,AF=AE,而AB=AF+FB,∴FB=BC,
仿前可证△BFD≌△BCD,得∠DCB=∠DFB……②
①、②两式中∵∠AFD+∠DFB=180°,∴∠AED+∠DCB=180°,于是AE∥BC.
2、回到原题,若D点在EB上,仍在AB上取AF=AE,连接DF、DC,套用上面的证法,得∠AED+∠DCB=180°,从而∠AEC+∠ECB>180°,则AE与BC必不平行.
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