Question

18．如图:已知正方形ABCD中，点E、F分别是BC、CD上，且AF平分∠DAE，则AE＝BE＋DF，请说明理由.

Answer 1

证明：假设正方形边长为a，∠DAF=∠FAE=θ
则有AE=AB/sin∠AEB=a/sin(2θ)
BE=AB/tan∠AEB=a/tan(2θ)
DF=AD/tan∠DAF=a/tanθ
BE+DF=a/tan(2θ)+a/tanθ
=a*[(1-tanθ^2)/(2tanθ)+1/tanθ]
=a*(1+tanθ^2)/(2tanθ)
=a*(sinθ^2+cosθ^2)/(2sinθ*cosθ)
=a/sin(2θ)
=AE