Junior Balkan MO

Solution:

Let us denote angles of triangle $ABC$ with $\alpha, \beta, \gamma$ in a standard way. By basic angle chasing we have
$$\angle BAD = 90^\circ - \beta = \angle OAC \text{ and } \angle CAD = \angle BAO = 90^\circ - \gamma$$
Using the fact that lines $AE$ and $AX$ are isogonal with respect to $\angle KAL$ we can conclude that $X$ is an $A$-antipode on $\omega$. (This fact can be purely angle-chased: we have
$$\angle KAX + \angle AXK = \angle KAX + \angle ALK = 90^\circ - \beta + \beta = 90^\circ$$
which implies $\angle AKX = 90^\circ$.) Now let $F$ be the projection of $X$ on the line $AE$. Using that $AX$ is a diameter of $\omega$ and $\angle EDX = 90^\circ$ it's clear that $F$ is the intersection point of $\omega$ and the circumcircle of triangle $DEX$. Now it suffices to show that $ABFC$ is cyclic. We have $\angle KLF = \angle KAF = 90^\circ - \gamma$ and from $\angle FEL = 90^\circ$ we have that $\angle EFL = \gamma = \angle ECL$ so quadrilateral $EFCL$ is cyclic. Next, we have
$$\angle AFC = \angle EFC = 180^\circ - \angle ELC = \angle ELA = \beta$$
(where last equality holds because of $\angle AEL = 90^\circ$ and $\angle EAL = 90^\circ - \beta$).

Solution:

We have $\angle BAD = 90^\circ - \beta = \angle OAC$ and that $AX$ is the diameter of $\omega$. Also we note that
$$\angle ALK = \beta, \quad \angle KLC = 180^\circ - \beta = \angle KBC$$
so $BKCL$ is cyclic. Let $AO$ intersect circumcircle of $ABC$ again at $A'$. We will show that $A'$ is the desired concurrence point. Obviously $AA'$ is the diameter of circumcircle of triangle $ABC$ so $\angle A'CA = 90^\circ$ which implies that $A'CLE$ is cyclic. From power of point $E$ we have that $EK \cdot EL = EB \cdot EC = EA \cdot EA'$, so we can conclude that $A' \in \omega$. Now using the fact that $AX$ is a diameter of $\omega$ implies $\angle AXA' = 90^\circ$ we have that $DXA'E$ is cyclic because of $\angle EDX = 90^\circ$ which finishes the proof. $\square$