Selection tests for the Balkan Mathematical Olympiad 2013

Applying Ptolemy relation to the cyclic quadrilateral $ABCD$, we get $$AB\cdot CD+BC\cdot DA=AC\cdot BD$$ which simplifies to $$CD+DA=25$$ Let $a=AB=BC=CA$ and $x=AE$. We have, from similarity of triangles $AED$ and $BEC$, that $$\frac{DA}{a}=\frac{x}{19}$$ We have, from similarity of triangles $CED$ and $BEA$, that $$\frac{CD}{a}=\frac{a-x}{19}.$$ We deduce that $$\frac{a}{19}=\frac{x}{19}+\frac{a-x}{19}=\frac{DA}{a}+\frac{CD}{a}=\frac{25}{a},$$ and therefore $a=5\sqrt{19}$. Applying sine law on the circumcircle of $ABCD$, we obtain $$\frac{\sin \angle BAD}{25}=\frac{\sin \angle BAC}{5\sqrt{19}}=\frac{\sqrt{57}}{190}.$$ Hence $\sin \angle BAD=\frac{5\sqrt{57}}{38}$, that is $$\cos \angle BAD=\pm \frac{\sqrt{38^2-25\times 57}}{38}=\pm \frac{\sqrt{19}}{38}.$$ Applying cosine law to triangle $ABD$, we obtain $$25^2=25\times 19+A{D}^2\pm 2\times 5\sqrt{19}\times AD\times \frac{\sqrt{19}}{38},$$ and therefore, $AD=10$ or $15$.