Selection Examinations for the IMO, 2013

The solution uses directed angles. Let $E$ be the midpoint of the side $AB$ and let $F$ be the midpoint of the side $AC$. Since $EF$ is a midline of the triangle $ABC$, $EF$ is parallel to $BC$. Because the lines $XY$ and $ZW$ are bisectors of the sides $AB$ and $AC$, respectively, the points $E, X$ and $Y$ as well as the points $F, Z$ and $W$ are collinear. The points $E, F, Y$ and $W$ are concyclic because
$$\angle WEY = \angle BEY = \frac{\pi}{2} = \angle WFC = \angle WFY.$$

From concyclicity of the points $E, F, Y$ and $W$, collinearity of the points $F, Z$ and $W$, and collinearity of the points $E, X$ and $Y$ we now derive
$$\angle XYW = \angle EYW = \angle EFW = \angle EFZ.$$
Further, from parallelism of the lines $EF$ and $BC$, collinearity of the points $F, W$ and $Z$, and collinearity of the points $B, C, X$ and $Y$ we get
$$\angle EFZ = \angle BZF = \angle XZF = \angle XZW.$$
The above two equalities say that $\angle XYW = \angle XZW$, hence the points $X, Y, Z$ and $W$ are concyclic.