Let $ABCD$ be a convex quadrilateral. Points $E$ and $F$ are midpoints of sides $\overline{AD}$ and $\overline{BC}$, respectively. Point $G$ is the intersection of lines $AF$ and $BE$, and point $H$ is the intersection of lines $DF$ and $CE$, respectively. Prove the area of quadrilateral $EHFG$ is equal to the sum of areas of triangles $\triangle AGB$ and $\triangle DHC$.
When $[ABC]$ is area of triangle $\Delta ABC$, then we have to show $$ [ABG] + [ CDH] =[EFG] + [EFH] $$
which is equivalent to $$ ([ABG] + [BFG])+([ CDH] +[CFH])=([EFG] + [BFG])+ ( [EFH] +[CFH] )\ (1)$$
Hence $A,\ E,\ D$ has a height $H,\ \frac{H+h}{2},\ h$ from a line segment $[BC]$, then $$ ([ABG] + [BFG])+([ CDH] +[CFH]) = \frac{1}{2} Hy +\frac{1}{2} hy\ (2) $$
$$([EFG] + [BFG])+ ( [EFH] +[CFH] ) = \frac{1}{2} \frac{H+h}{2}y + \frac{1}{2} \frac{H+h}{2}y\ (3) $$ Hence $(2),\ (3)$ gives $(1)$