22nd Junior Turkish Mathematical Olympiad

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