Estonian Mathematical Olympiad

As $\frac{MY}{MB}=\frac{2ME}{2ME}=1$ and analogously $\frac{MZ}{MC}=1$ we have $BC\parallel YZ$. Let $G$ be the intersection of lines $YZ$ and $AD$ (Fig. 14).

Then $\frac{MG}{MD}=\frac{MY}{MB}=1$ from which $MG=MD=\frac{1}{3}AD$ and $AG=AD-MD-MG=\frac{1}{3}AD$. Hence $\frac{AL}{AB}=\frac{AG}{AD}=\frac{1}{3}$.

By swapping the roles of $A$ and $B$, the roles of $D$ and $E$, roles of $X$ and $Y$, and finally the roles of $K$ and $L$, we get analogously $\frac{BK}{AB}=\frac{1}{3}$. Therefore $AL=BK$.

Fig. 14

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Alternative solution.

Notice that triangle $XYZ$ is a homothetic transformation of triangle $ABC$ with centre $M$ and ratio $-1$. Homothety preserves the directions of the lines, therefore $KL\parallel XY$ and $F$ lies on the median drawn from vertex $Z$ of triangle $XYZ$. Thus $FK=FL$ and $AL=AF-FL=BF-FK=BK$.