如图,在三角形ABC中,AB=AC,角BAC=90度,D为AC终点,AE垂直于F交BC于E求证角ADB=角CDE
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如图,在三角形ABC中,AB=AC,角BAC=90度,D为AC终点,AE垂直于F交BC于E求证角ADB=角CDE
用等边三角形
![](http://img.wesiedu.com/upload/f/02/f029413134649a43edba80c36e254a9e.jpg)
用等边三角形
![](http://img.wesiedu.com/upload/f/02/f029413134649a43edba80c36e254a9e.jpg)
![如图,在三角形ABC中,AB=AC,角BAC=90度,D为AC终点,AE垂直于F交BC于E求证角ADB=角CDE](/uploads/image/z/6201100-28-0.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAB%3DAC%2C%E8%A7%92BAC%3D90%E5%BA%A6%2CD%E4%B8%BAAC%E7%BB%88%E7%82%B9%2CAE%E5%9E%82%E7%9B%B4%E4%BA%8EF%E4%BA%A4BC%E4%BA%8EE%E6%B1%82%E8%AF%81%E8%A7%92ADB%3D%E8%A7%92CDE)
从C点作CG‖AB,与AE的延长线交于G,〈BAC+〈GCA=180度,〈BAD=〈ACG=90度,AC=AB,AF⊥BD,〈FAD+〈ADF=90度,〈DBA+〈FDA=90度,〈CAG=〈ABD,
△ACG≌△BAD,〈CGA=〈ADB,AD=CG(对应边相等),D为AC中点,AD=CD,CD=CG,〈ECG=〈ABD=45度(内错角相等),〈DCE=45度,CE=CE(公用边),
△CDE≌△CGE,<CGE=<CDE,前已证明〈CGE(A)=〈ADB,
∴〈ADB=〈CDE
![](http://img.wesiedu.com/upload/d/3b/d3b90a9bbc70f668a9848a9dad63b8b5.jpg)
△ACG≌△BAD,〈CGA=〈ADB,AD=CG(对应边相等),D为AC中点,AD=CD,CD=CG,〈ECG=〈ABD=45度(内错角相等),〈DCE=45度,CE=CE(公用边),
△CDE≌△CGE,<CGE=<CDE,前已证明〈CGE(A)=〈ADB,
∴〈ADB=〈CDE
![](http://img.wesiedu.com/upload/d/3b/d3b90a9bbc70f668a9848a9dad63b8b5.jpg)
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