Belarusian Mathematical Olympiad

Let $F^{\prime }$ be the foot of an altitude from $A$ in the triangle $ABC$.



It is enough to prove that the points $D$, $F^{\prime }$ and $E$ are colinear, since it will lead to $F=F^{\prime }$ which means that $AF$ is an altitude of the triangle $ABC$. Since $\angle CDA=\angle C{F}^{\prime }A=90^{\circ }$, the quadrilateral $CD{F}^{\prime }A$ is cyclic and therefore $\angle C{F}^{\prime }D=\angle CAD$. Since $\angle A{F}^{\prime }L=\angle AEL=90^{\circ }$, the quadrilateral $AE{F}^{\prime }L$ is cyclic and hence $\angle BE{F}^{\prime }=\angle LAE$. Since $\angle CAL=\angle LAB$, then $\angle C{F}^{\prime }D=\angle B{F}^{\prime }E$ which means that $D$, $F^{\prime }$ and $E$ are colinear.

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Alternative solution.

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 $CL$.