Let $M,N,P,Q,R$ be the midpoints of $AC, AB, BE, CD, BC$. Clearly $RPN, RQM$ are collinear. Note by similarity that $RP \cdot RN = \frac{1}{4} CE \cdot CA = \frac{1}{4} BC^2$, since $(BEA)$ is tangent to $BC$. Similarly we deduce $RQ*RM = \frac{1}{4} BC^2 \implies RP\cdot RN = RQ\cdot RM \implies P,Q,M,N \text{ are concyclic.}$.

Dear Mathlinkers, a proof without any calculation is possible... I am waiting for your suggestion... Sincerely Jean-Louis

Dear Mathlinkers, any ideas? Sincerely Jean-Louis

This problem was proposed by me

Let $M,N,K,L,P$ be midpoints of $AB,AC,BE,CD,BC$ . $\angle BMK = \angle BAC = \angle BMP$ ---> $M,K,P$ are collinear. $\angle CNL = \angle BAC = \angle CNP$ ---> $N,L,P$ are collinear. $\angle BMK = \angle PBK$ ---> $BP$ is tangent to $BMK$ circumcircle ---> $PB^2 = PK.PM$ $\angle CNL = \angle LCP$ ---> $CP$ is tangent to $CNL$ circumcircle ---> $PC^2 = PL.PN$ $PB = PC$ ---> $PK.PM$ = $PL.PN$ ---> $NMKL$ is cyclic.

Let $F,K,L,M,N$ be the midpoints of $\overline{BC},\overline{AB},\overline{AC},\overline{BE},\overline{CD},$ respectively and notice $\triangle CFN\sim\triangle LFC$ and $\triangle KBF\sim\triangle BMF.$ Hence, $$FN\cdot FL=CF^2=BF^2=FM\cdot FK.$$$\square$