31st Hellenic Mathematical Olympiad

We extend $AM$ till it meets $ED$ at point $N$. Then $BMNE$ and $MCDN$ are parallelograms and hence $EN=BM=MC=ND$. Hence $N$ is the middle of $ED$. Moreover we observe that $\frac{AM}{MN}=2$ and $M$ lie on the median of the triangle $EAD$. Hence $M$ is the centroid of the triangle $AED$. Therefore the line $EM$ is the line of the median of the triangle $AED$ passing from the vertex $E$, and so it intersects the side $AD$ in the middle.