几何题超难如图http://bar.hexun.com/d/385643.html ;已知在△ABC中,AC=BC,CD⊥AB,垂足为D,E、G为CD上两点,且∠1=∠2=∠3,AF交CD于G,交BC于F,连接EF,则GB‖EF,请说明理由.B点处是∠1=∠2=∠3

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几何题超难如图http://bar.hexun.com/d/385643.html ;已知在△ABC中,AC=BC,CD⊥AB,垂足为D,E、G为CD上两点,且∠1=∠2=∠3,AF交CD于G,交BC于F,连接EF,则GB‖EF,请说明理由.B点处是∠1=∠2=∠3

几何题超难如图http://bar.hexun.com/d/385643.html ;已知在△ABC中,AC=BC,CD⊥AB,垂足为D,E、G为CD上两点,且∠1=∠2=∠3,AF交CD于G,交BC于F,连接EF,则GB‖EF,请说明理由.B点处是∠1=∠2=∠3
几何题超难
如图http://bar.hexun.com/d/385643.html ;已知在△ABC中,AC=BC,CD⊥AB,垂足为D,E、G为CD上两点,且∠1=∠2=∠3,AF交CD于G,交BC于F,连接EF,则GB‖EF,请说明理由.
B点处是∠1=∠2=∠3

几何题超难如图http://bar.hexun.com/d/385643.html ;已知在△ABC中,AC=BC,CD⊥AB,垂足为D,E、G为CD上两点,且∠1=∠2=∠3,AF交CD于G,交BC于F,连接EF,则GB‖EF,请说明理由.B点处是∠1=∠2=∠3
连结 AE
因为是等腰三角形,所以有：∠CAE=∠EAG=∠4=∠1
记AF和EB交点为O
则由 ∠EAO=∠1,∠EOA=∠FOB,知△AEO和△BFO相似
所以有：EO:AO=FO:BO
又有：∠EOF=∠AOB
得到 △EOF和△AOB相似
所以 ∠FEB=∠1=∠2
内错角相等,得GB‖EF

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