延长GH交AD于H,延长FE交DC于I,设HE=a,EI=b.由该四边形为正方形得:
AF=FE=a,EG=EI=b.则:DE^2=FG^2=a^2+b^2.故:DE=FG
设DE=FG=c,延长DE交FG或延长线于P,EP=ab/c则:EP/HE=(ab/c)/a=b/c
EG/ED=b/c,则:ΔHED∽ΔPEG,则∠EPG=90°,即:DE⊥FG
延长GH交AD于H,延长FE交DC于I,设HE=a,EI=b.由该四边形为正方形得:
AF=FE=a,EG=EI=b.则:DE^2=FG^2=a^2+b^2.故:DE=FG
设DE=FG=c,延长DE交FG或延长线于P,EP=ab/c则:EP/HE=(ab/c)/a=b/c
EG/ED=b/c,则:ΔHED∽ΔPEG,则∠EPG=90°,即:DE⊥FG