如图,CD⊥AB,EF⊥AB,垂足分别为点D,F,CD=EF,AE=BC,求证:AD=BF
人气:116 ℃ 时间:2020-04-23 14:06:01
解答
CD⊥AB,EF⊥AB,垂足分别为点D,F,CD=EF,AE=BC,
所以EFDC是矩形,CE=DF,
直角三角形AEF全等BCD,
所以,AF=DB,
AF-DF=BD-DF,
所以,AD=BF.
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