Given a quadrilateral $ABCDEF$, where $B$ is the midpoint of $\overline{AC}$ and $E$ is the midpoint of $\overline{FD}$, prove the relation $$a+d = c+b,$$ where $a$, $b$, $c$, $d$ denote the respective areas of triangle as denoted by the following figure:
Let $O_1$ be $AE$ intersect $BF$, $O_2$ be $BD$ intersect $CE$.
$$S_{FO_1A}+S_{DO_2C}=S_{BO_1E}+S_{BO_2E} \Leftrightarrow$$ $$S_{FO_1A}+S_{O_1AB}+S_{DO_2C}+S_{O_2BC}=S_{BO_1E}+S_{O_1AB}+S_{BO_2E}+S_{O_2BC} \Leftrightarrow$$ $$ S_{FAB}+S_{DBC}=S_{AEC}$$
Let $FH_1, EH_2, DH_3$ be perpendiculars, dropped onto line $AC$. Then we need to prove only $FH_1+DH_3=2EH_2$ and it holds because of $EH_2$ being midline of $FH_1H_3D$.