Belarusian Mathematical Olympiad

Let $F'$ be the foot of an altitude from $A$ in the triangle $ABC$.

It is enough to prove that the points $D$, $F'$ and $E$ are colinear, since it will lead to $F = F'$ which means that $AF$ is an altitude of the triangle $ABC$. Since $\angle CDA = \angle CF'A = 90^\circ$, the quadrilateral $CDF'A$ is cyclic and therefore $\angle CF'D = \angle CAD$. Since $\angle AF'L = \angle AEL = 90^\circ$, the quadrilateral $AEF'L$ is cyclic and hence $\angle BEF' = \angle LAE$. Since $\angle CAL = \angle LAB$, then $\angle CF'D = \angle BF'E$ which means that $D$, $F'$ and $E$ are colinear.

Let the lines $EL$ and $CD$ meet at $K$. Since the angles $ADK$ and $AEK$ are right, the points $A$, $K$, $D$ and $E$ lie on the circle with diameter $AK$. Hence $\angle DKE = \angle DAE$. Since $AD$ is the bisector, $\angle DAE = \angle DAC$, whence $\angle DKE = \angle DAC$. The latter is equivalent to $\angle DKL = \angle CAL$, whence $A$, $L$, $C$ and $K$ lie on a circle.
Consider the Simson line of $A$ and the triangle $LCK$. The foot of the perpendicular from $A$ to $CK$ is $D$, and to $LK$ is $E$, therefore this line is $DE$. Since $DE$ meet $CL$ at $F$, $F$ is the foot of the perpendicular from $A$ to $CL$.