Question

如图,在矩形abcd中,E为AD中点,EF垂直EC交AB于F,联结FC（AB＞AE）.△AEF与△EFC相似吗?请说明理

Answer 1

∠AEF = 180°-∠CEF-∠CED = 90°-∠CED = ∠DCE ,

因为,∠EAF = 90°= ∠CDE ,∠AEF = ∠DCE ,

所以,△AEF ∽ △DCE ,

可得：EF∶CE = AF∶DE = AF∶AE ；

因为,∠EAF = 90°= ∠CEF ,EF∶CE = AF∶AE ,

所以,△AFE ∽ △EFC .