Let $A B C D$ be a parallelogram. Let $E$ be a point such that $A E=B D$ and $(A E)$ is parallel to $(B D)$. Let $F$ be a point such that $E F=A C$ and $(E F)$ is parallel to $(A C)$. Express the area of the quadrilateral $A E F C$ in terms of the area of $A B C D$.
As $A E / / D B$ and $A E = D B$, $A E B D$ is a parallelogram. Therefore, $E B = A D = B C$ and $E B / / A D / / B C$, in other words, $B$ is the midpoint of segment $E C$. Furthermore, since $E F = A C$ and $E F / / A C$, $A E F C$ is a parallelogram, and thus $B$ is the point of intersection of its diagonals. Therefore, the triangles $C D A$, $A B C$, and $F B E$ are isometric, and thus have equal areas.

We can similarly show that $A E B$, $B D A$, $D B C$, and $F C B$ are isometric, and therefore also have equal areas.

Thus, the area of $A E F C$ is twice the area of $A B C D$.