0、谁会卡www.88cfcf.com点?、bC膊2 N9 2

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访问统计:已知正方形ABCD,点E在CD上,且CE:DE=1:2,EF⊥EA,交与点F,则EF:EA?_百度知道
已知正方形ABCD,点E在CD上,且CE:DE=1:2,EF⊥EA,交与点F,则EF:EA?
提问鍺采纳
设AD=CD=3a∵CE:DE=1:2∴CE=1/3CD=aDE=2/3CD=2a∵EF⊥EA∴∠DAE=∠CEF(同为∠AED的余角)∴Rt△ADE∽Rt△ECF∴AD/CE=DE/CFCF=CE×DE/AD=a×2a/3a=2/3a∴在Rt△CEF中EF=√[CE²+CF²)=√[a²+(2/3a)²=√13/3a在Rt△ADE中:EA²=AD²+DE²=(3a)²+(2a)²=13a²EA=√13a∴EF∶EA=(√13/3)a∶(√13)a=1∶3
提问者评价
谢谢你的耐心解答,好详细吖
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其他2条回答
题中&EF⊥EA,交与点F&,是点F在BC边上吗?如是,计算如下:∵△ADE∽△ECF∴AD/CE=DE/CF∵DE=2AD/3  CE=AD/3 CF=DE*CE/AD=2AD/9∵EA^2=AD^2+DE^2=(13/9)AD^2 EF^2=CE^2+CF^2=(13/81)AD^2∴EF/EA=1/3
EAD-EFC相似EF:EA=EC:AD = 1/3
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