看圖令正方形的邊長為a,角DAF=α,角EAF=α,則角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,原命題得證
注釋：^2表示平方,考慮到α和2α均為銳角,所以對它們正弦、余弦等的開方均為正值