Suppose Bernd chooses $Y$ such that $XY$ is parallel to $AB$. If $\frac{BX}{BC}=k$, then ratio of the height from $Z$ to $XY$ to the height from $C$ to $AB$ is also $k$, and the ratio $\frac{XY}{AB}=1-k$. Both of these can be found by using similar triangles. So the ratio $\frac{[XYZ]}{[ABC]}=k(1-k)$, which means $[XYZ]=k(1-k)$. Obviously, the maximum value of this is $\frac{1}{4}$, so Anja cannot create a triangle greater than $\frac{1}{4}$ if Bernd plays optimally. Now suppose Anja chooses $X$ to be the midpoint of $BC$. Then if Bernd chooses $Y$ to be closer to $C$ than $A$, then should Anja should choose $Z=A$. This will create a triangle with area greater than $\frac{1}{4}$ since the height from $X$ to $YZ$ is exactly half of the height from $B$ to $AC$ and also $YZ>\frac{AC}{2}$. If Bernd chooses $Y$ to be closer to $A$ than $C$, then Anja should choose $Z=B$. This will also create a triangle with area greater than $\frac{1}{4}$ since $XZ$ is exactly half of $BC$ but the height from $Y$ to $BC$ is more than half of the height from $A$ to $BC$. Therefore, Bernd should choose $Y$ to be the midpoint of $AC$, which forces the triangle to have area $\boxed{\frac{1}{4}}$ regardless of $Z$, since we know $XY$ is parallel to $AB$.