Source: Challenge And Thrill Of Pre College Mathematics
On the line segment $AB$, parallelograms $ABCD$ and $ABXY$ are drawn on opposite sides and $AB$ or $AB$ produced bisects $CX$. Prove that $ABCD$ and $ABXY$ have equal area.
Suppose $AB$ intersects $CX$ at $M$. Then if $\triangle BMC$ can be proven congruent to $\triangle BMX$ then that would mean that their heights are $MC = MX$. I am unable to proceed further from here.
Since the line $AB$ bisects the interval $CX$, the perpendiculars dropped from $C$ and from $X$ to the line $AB$ have the same length. This means that the parallelograms $ABCD$ and $ABXY$ have the same altitude. Since they have the same base, it follows that the areas are the same.