Question

如图,在正方形ABCD中,E、F分别在BC、CD上,∠EAF=45°,试证明S△AEF=S△ABE=S△ADF

Answer 1

试证明S△AEF=S△ABE=S△ADF。打错！ 应该是：试证明S△AEF=S△ABE+S△ADF。

把⊿ABE绕A逆时针旋转90°，到达⊿ADG.

∠FAG＝∠FAD+∠DAG＝∠FAD+∠BAE＝90º-45º＝15º＝∠FAE

又AG＝AE,AF＝AF,∴⊿AFG≌⊿AFE﹙SAS﹚.

S⊿AFE＝S⊿AFG＝S⊿AFD+S⊿ADG＝S⊿AFD+S⊿ABE