Saudi Arabian Mathematical Competitions

First, we will show that $D$ is the orthocenter of triangle $ABC$. Denote $A' = AD \cap BC$, $B' = BD \cap CA$, $C' = CD \cap AB$. Since $\angle DAB = \angle DCB$, we have $ACA'C'$ is cyclic. Similarly, $ABA'B'$ is also cyclic. So we have
$$\angle DA'B = \angle DB'A, \quad \angle DA'C = \angle DC'A \text{ and } DA \cdot DA' = DB \cdot DB' = DC \cdot DC',$$
so $BCB'C'$ is also cyclic. Thus $\angle DB'A = \angle DC'A$ which implies that $\angle DA'B = \angle DA'C$, but $\angle DA'B + \angle DA'C = 180^\circ \rightarrow AA' \perp BC$. It leads to $BB' \perp CA$, $CC' \perp AB$ then $D$ is the orthocenter of triangle $ABC$.

Continue, denote $M$ as the intersection of $AE$ and $BC$ then $\angle EBM = \angle EAB$ implies that $MB$ is tangent to $(ABE)$. Similarly, $MB$ is tangent to $(ACE)$ then
$$ME \cdot MA = MB^2 = MC^2 = MB'^2 = MC^2$$
(since $MB = MC = MB' = MC'$). So $MB'$ is tangent to $(AEB')$, thus $\angle AEB' = \angle MB'C = \angle C$. Similarly, $\angle AEC' = \angle B$ then
$$\angle B'EC' = \angle AEB' + \angle AEC' = \angle B + \angle C = 180^\circ - \angle A.$$
Thus $AB'EC'$ is cyclic, but $AB'DC'$ is cyclic, so five points $A, B', C', D, E$ are concyclic, which implies that
$$\angle AED = \angle AB'D = \angle AC'D = 90^\circ.$$
Therefore, $ADE$ is a right triangle. $\square$