Let $\Delta \mathrm\,{ABC}$ be a triangle in the plane and $X,\, Y,\,Z$ be points on sides $BC,\, CA,\,AB$, respectively.
If lines $XY$ and $AB$ are not parallel, there is a location for $Z$ for each location of points $X$ and $Y$ so that the area of $\Delta\,XYZ$ becomes maximal.
Prove that, if lines $XY$ and $AB$ are parallel, the area of $\Delta\,XYZ$ is smaller than every such possible maximum area of $\Delta\,XYZ$ for non-parallel $XY$ and $AB$.
I have already figured that, for given $X$ and $Y$, the area of $\Delta\,XYZ$ becomes maximal if $Z$ lies on that ending of segment $AB$ which has a larger distance to line $XY$ (so $Z$ is located on either point $A$ or point $B$).
Also, when $XY$ an $AB$ are parallel, there is only one possible area for $\Delta\,XYZ$ as the perpendicular height and hence the area are constant and don't depend on the location of $Z$.
But how does that help in solving the actual task?
I am grateful for any hints on this!