THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Let $N$ be the midpoint of $[AD]$; it follows that $BN\perp AD$. If $AP\perp BE$, $P\in BE$; then the quadrilateral $BNAP$ is a rectangle.

Since the points $N$ and $P$ are the projections of $A$ on the interior and exterior bisectors of the angle $ABC$, respectively, it follows that the line $NP$ contains the midline parallel to $BC$, and hence $M\in (NP)$.

Let $\{F\}=AC\cap BE$; from $△BAE\sim △BCF$ it follows that $\angle AEB=\angle BFC$ and $\angle AFE=\angle AEB$, hence the triangle $AEF$ is isosceles, with $AE=AF$, and we deduce that $P$ is the midpoint of the line segment $[EF]$.

In the trapezoid $ADFE$, the points $N$ and $P$ are the midpoints of the bases, therefore the line $NP$ contains the point of intersection of trapezoid's diagonals $[AF]$ and $[DE]$. Since $\{M\}=AC\cap NP$, it follows that $M\in DE$.