Notes
triangles with a parallelogram solution

Triangles with a Parallelogram

Solution by Similar Triangles and Area Scale Factor

With the points labelled as above, the area of the parallelogram can be found by multiplying the lengths of $F G$ and of $I H$.

The line segments $F C$ and $A E$ are parallel, and angles $F \hat{B} C$ and $E \hat{B} A$ are vertically opposite and so are equal. Therefore, triangles $A B E$ and $F B C$ are similar. The area scale factor is $9$, so the length scale factor is $3$. This means that the length of $I H$ is $\frac{4}{3}$ times that of $B I$.

A similar argument applied to triangles $F C B$ and $G C D$ shows that $G F$ is $\frac{5}{3}$ of $F C$.

Since the area of triangle $F B C$ is $45$, multiplying the lengths of $F C$ and $B I$ gives $90$. Therefore the area of the parallelogram is $\frac{4}{3} \times \frac{5}{3} \times 90 = 200$.