-(>口<-)「初二数学提问.」-如图,在△ABC中,∠ACB=90°,CE⊥AB于点E,AD=AC,AF平分∠CAB交CE于点F,DF的延长线交AC于点G,求证:⒈DF‖BC⒉FG=FE.http://hiphotos.baidu.com/%5F%9A%EC%5F/pic/item/210f36829d7fa49ff603a6c3.jpg
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 10:08:20
![-(>口<-)「初二数学提问.」-如图,在△ABC中,∠ACB=90°,CE⊥AB于点E,AD=AC,AF平分∠CAB交CE于点F,DF的延长线交AC于点G,求证:⒈DF‖BC⒉FG=FE.http://hiphotos.baidu.com/%5F%9A%EC%5F/pic/item/210f36829d7fa49ff603a6c3.jpg](/uploads/image/z/10450881-9-1.jpg?t=%EF%BC%8D%28%3E%E5%8F%A3%EF%BC%9C-%29%E3%80%8C%E5%88%9D%E4%BA%8C%E6%95%B0%E5%AD%A6%E6%8F%90%E9%97%AE.%E3%80%8D-%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ACB%3D90%C2%B0%2CCE%E2%8A%A5AB%E4%BA%8E%E7%82%B9E%2CAD%3DAC%2CAF%E5%B9%B3%E5%88%86%E2%88%A0CAB%E4%BA%A4CE%E4%BA%8E%E7%82%B9F%2CDF%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4AC%E4%BA%8E%E7%82%B9G%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%92%88DF%E2%80%96BC%E2%92%89FG%3DFE.http%3A%2F%2Fhiphotos.baidu.com%2F%255F%259A%25EC%255F%2Fpic%2Fitem%2F210f36829d7fa49ff603a6c3.jpg)
-(>口<-)「初二数学提问.」-如图,在△ABC中,∠ACB=90°,CE⊥AB于点E,AD=AC,AF平分∠CAB交CE于点F,DF的延长线交AC于点G,求证:⒈DF‖BC⒉FG=FE.http://hiphotos.baidu.com/%5F%9A%EC%5F/pic/item/210f36829d7fa49ff603a6c3.jpg
-(>口<-)「初二数学提问.」
-如图,在△ABC中,∠ACB=90°,CE⊥AB于点E,AD=AC,AF平分∠CAB交CE于点F,DF的延长线交AC于点G,求证:⒈DF‖BC⒉FG=FE.
http://hiphotos.baidu.com/%5F%9A%EC%5F/pic/item/210f36829d7fa49ff603a6c3.jpg
-(>口<-)「初二数学提问.」-如图,在△ABC中,∠ACB=90°,CE⊥AB于点E,AD=AC,AF平分∠CAB交CE于点F,DF的延长线交AC于点G,求证:⒈DF‖BC⒉FG=FE.http://hiphotos.baidu.com/%5F%9A%EC%5F/pic/item/210f36829d7fa49ff603a6c3.jpg
方法:此题要掌握基本图形:直角三角形斜边上的高的图形,会得出很多条件
1)三角形ACF全等于三角形ADF(SAS)[AC=AD,角CAF=角DAF,AF=AF],所以角ACF=角ADF
因为角ACF+角ECB=90度,角ECB+角B=90度,所以角ACF=角B,所以角ADF=角B
所以DF平行BC
2)因为DF平行BC,所以AGD=角ACB=90度.因为AF平分∠CAB,所以角CAF=角DAF.所以三角形AGF全等于三角形AEF(AAS),所以FG=FE