Lời giải:
Ta có:
$\quad \widehat{AMC}=\dfrac12\left(sđ\mathop{AC}\limits^{\displaystyle\frown} + sđ\mathop{BD}\limits^{\displaystyle\frown}\right)$
$\Leftrightarrow \widehat{AMC}= \dfrac12\left(sđ\mathop{AB}\limits^{\displaystyle\frown} + sđ\mathop{BD}\limits^{\displaystyle\frown}\right)$
$\Leftrightarrow \widehat{AMC}= \dfrac12sđ\mathop{AD}\limits^{\displaystyle\frown}$
Ta lại có:
$\widehat{AED}= \dfrac12sđ\mathop{AD}\limits^{\displaystyle\frown}$ (góc nội tiếp chắn $\mathop{AD}\limits^{\displaystyle\frown}$)
Do đó:
$\widehat{AMC}=\widehat{AED}$
Hay $\widehat{AMN}=\widehat{NED}$
Xét tứ giác $DMNE$ có:
$\widehat{AMN}=\widehat{NED}\quad (cmt)$
Do đó $DMNE$ là tứ giác nội tiếp
