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
$$25^2 = 25 \times 19 + AD^2 \pm 2 \times 5\sqrt{19} \times AD \times \frac{\sqrt{19}}{38},$$
and therefore, $AD = 10$ or $15$.