如图5,菱形ABCD中,∠B=60°,点E在边BC上,点F在边CD上,∠AEF=60°,求证...答:菱形ABCD中,∠B=60°,点E在边BC上,点F在边CD上,∠AEF=60°,设AB=1,BE=x,0<x<1,∠BAE=a,则∠CEF=120°-∠AEB=∠BAE=a,∠CFE=60°-a,EC=1-x,由正弦定理,AB/sin(60°+a)=BE/sina,∴sina=xsin(60°+a),① AE=BEsinB/sina=x√3/(2sina),EF=CEsinC/sin∠CFE=(1-...