24th Balkan Mathematical Olympiad

Let $\angle BAC=\angle BCA=\alpha$, $\angle CBD=\angle CDB=\beta$, and $S$ the point of intersection of the lines $AB$ and $DC$. If $AE=DE$, then $\frac{AE}{\sin (2\alpha +\beta )}=\frac{AB}{\sin (\alpha +\beta )}=\frac{CD}{\sin (\alpha +\beta )}=\frac{DE}{\sin (\alpha +2\beta )}$.

Now is obvious that $$\sin (2\alpha +\beta )=\sin (\alpha +2\beta ),0^{\circ }<2\alpha +\beta <180^{\circ },0^{\circ }<\alpha +2\beta <180^{\circ },$$ $\alpha \ne \beta$ (since $AC\ne BD$), and therefore $$2\alpha +\beta +\alpha +2\beta =180^{\circ }\text{ or }\alpha +\beta =60^{\circ }.$$ Hence we have: $\angle BAD+\angle ADC=120^{\circ }$.

Conversely, if $\angle BAD+\angle ADC=120^{\circ }$ we have $$\angle AEB=\angle BAE+\angle EDC(1)$$ $$\angle AEB=\angle EAD+\angle EDA(2)$$ From (1) and (2) we have $$2(\angle AEB)=\angle BAD+\angle ADC=120^{\circ }.$$ Hence $\angle AEB=60^{\circ }=\angle BSC$ and the quadrilateral $SBEC$ is inscribable. Therefore $\angle BSE=\angle BCE=\angle BAE$ and $AE=ES$. Similarly we prove that $ES=ED$.