Junior Turkish Mathematical Olympiad

Using Stewart's Theorem in triangles $ABD$ and $BCD$, we obtain
$$\frac{AB^2 \cdot ED + AD^2 \cdot BE}{BD} - \frac{BE \cdot ED}{AE^2} = \frac{BC^2 \cdot ED + CD^2 \cdot BE}{BD} - \frac{BE \cdot ED}{EC^2}.$$
Using the given relations
$$AB^2 \cdot ED = CD^2 \cdot BE \quad \text{and} \quad AD^2 \cdot BE = BC^2 \cdot ED$$
we get $AE = EC$. Using the first equation, we obtain
$$\frac{BE}{BD}(AD^2 + CD^2) = AE^2 + BE \cdot ED$$
By the median length equation in triangle $ADC$, we obtain
$$AD^2 + CD^2 = 2(AE^2 + ED^2)$$
and hence
$$\left( 2 \cdot \frac{BE}{BD} - 1 \right) AE^2 = BE \cdot ED - 2 \cdot ED^2 \cdot \frac{BE}{BD}$$
The last equation is equivalent to
$$(AE^2 - BE \cdot ED) \left( \frac{BE - ED}{BD} \right) = 0$$
Therefore we have $AE^2 = BE \cdot ED$ or $BE = ED$. In the first case $ABCD$ is a cyclic quadrilateral and in the second case it is a parallelogram.