Ptolemy's on $ABCD$ gives $DC + DA = 25$. Then, $\triangle AED \sim \triangle BEC$, and $\triangle AEB \sim \triangle DEC$, implying that $$\frac{CD}{AB} = \frac{AB-AE}{19}$$$$\implies AB = BC = CA = 5 \sqrt{19}$$Law of Sines on $ABCD$ gives $$\frac{\sin \angle BAD}{25} = \frac{\sin \angle BAC}{5 \sqrt{19}} = \frac{\sqrt{57}}{190}$$$$\implies \cos \angle BAD = \frac{\sqrt{19}}{38}$$And finally Law of Cosines on $\triangle ABD$ gives $AD = 10$ or $AD = 15$ $\quad \blacksquare$