Ta có:
$M$ là điểm chính giữa $\mathop{AB}\limits^{\displaystyle\frown}$
$\to \mathop{AM}\limits^{\displaystyle\frown}=\mathop{BM}\limits^{\displaystyle\frown}$
Ta lại có:
$\widehat{MEB}= \dfrac12(s₫\mathop{BM}\limits^{\displaystyle\frown}+ s₫\mathop{AD}\limits^{\displaystyle\frown})$
$\to \widehat{MEB}=\dfrac12(s₫\mathop{AM}\limits^{\displaystyle\frown}+ s₫\mathop{AD}\limits^{\displaystyle\frown})$
$\to \widehat{MEB}=\widehat{MEP}=s₫\dfrac12\mathop{MD}\limits^{\displaystyle\frown}$
Mặt khác:
$\widehat{MCD}=\widehat{PCD}=\dfrac12s₫\mathop{MD}\limits^{\displaystyle\frown}$
Do đó:
$\widehat{MEP}=\widehat{PCD}$
Xét tứ giác $DEPC$ có:
$\widehat{MEP}=\widehat{PCD}\quad (cmt)$
Do đó $DEPC$ là tứ giác nội tiếp
