Let $B'$ and $D'$ on $AC$ be the feet of the altitudes of the triangles $\triangle ABC$ and $\triangle ADC$. Because these two triangles have the same area, we get $|BB'| = |DD'|$. This implies that the two right triangles $BMB'$ and $DMD'$, which have equal angles at $M$, are congruent. In particular, $|BM| = |MD|$.

A similar argument, using the other diagonal, shows that $|AM| = |MC|$. Because the two triangles $DMA$ and $BMC$ have equal angles at $M$, they are now seen to be congruent. This shows that $\angle MBC = \angle MDA$, hence $BC$ is parallel to $DA$. In a similar way it follows that $AB$ is parallel to $CD$, hence $ABCD$ is a parallelogram.