在△ABC中,∠ACB=90°,AC=BC,D是AB上一点,AE⊥CD于E,BF⊥CD交CD的延长线于F,CH⊥AB于H,CH交AE于G.求证:BD=CG./%CE%FB%CE%FB%BA%C7%BA%C7%B9%FE%B9%FE/pic/item/1e7cbfeecf1b3e1a.jpg图在这,谢谢
貌似F这个点没用啊...证明三角形CDH全等于三角形AGH.(CH=AH,角CDH=角CGE=角AGH),则HG=HD又CH=HB所以HB-HD=CH-CG既CG=BD
为您推荐:
请你发个图来好吗没有图我不太会做
LZ乃很好很强大直角三角形斜边大于直角边,怎么可能等于
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