Question

已知点E、F分别在正方形ABCD的边BC、CD上,且∠DAF=∠EAF.求证:AE=BE+DF

Answer 1

看图令正方形的边长为a,角DAF=α,角EAF=α,则角AEB=2α

由于tagα=DF/AD,所以DF=AD*tagα=atagα

tag2α=AB/BE,所以BE=AB/tag2α=a/tag2α

BE+DF=a/tag2α+a*tagα

=a*(1-tagα^2)/(2*tagα)+a*tagα

=a*(1-tagα^2)/(2*tagα)+a*2tagα^2/(2*tagα)

=a*(1+tagα^2)/(2*tagα)

=a/(2*tagα*cosα^2)

=a/sin(2α)

AE^2=AB^2+BE^2

=a^2+a^2/tag(2α)^2

=a^2[1+1/tag(2α)^2]

=a^2*(sec2α)^2/tag(2α)^2

=a^2/sin(2α)^2

所以AE=a/sin(2α)

因此,BE+DF=AE,原命题得证

注释：^2表示平方,考虑到α和2α均为锐角,所以对它们正弦、余弦等的开方均为正值