Question

如图，△ABC中，∠B=90°，∠C=45°，AB=3，D是BC上的点，且CD=1

如图，△ABC中，∠B=90°，∠C=45°，AB=3，D是BC上的点，且CD=1，线段AD的垂直平分线与AB，AD分别相交于点E，F。
求：线段AE的长。

Answer 1

∵∠A=∠C=45°（你知道的）
∴AB=BC=3
∴BD=BC-DC=3-1=2
再根据勾股定理求出AD
∴AD=根号13
∴AF=2分之根号13
∵∠AEF=∠B=90°
∴△AEF∽△ABD
∴AE/AD=AF/AB
∴AE/根号13=2分之根号13/3
∴AE=6分之13
说得够明白了吧

Answer 2

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Answer 3

先连接ED两点，因为C角为45度，CD=1,AB=3.则BD=2;很显然用两三角型相似定理。三角形AEF与三角形ABD相似。即AB/AD=AF/AE得。AE=13/6。