In the following figure-not drawn to scale!- $E$ is the midpoint of $B C$, triangle $F E C$ has area 7, and quadrilateral $D B E G$ has area 27. Triangles $A D G$ and $G E F$ have the same area, $x$. Find $x$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6cb05c45fe7c141f3316g-4.jpg?height=661&width=700&top_left_y=1209&top_left_x=688)
The answer is $x=8$.

Use the notation [.] to denote the area of a polygon. Draw $G B$; notice that triangles $G B E$ and $G E C$ have equal bases and altitudes, so $[G B E]=[G E C]=x+7$. Since $[A B E]=27+x$, we have $[G D B]=$ $20-x$.

Likewise, if we draw $A C$, we see that $[A B E]=[A E C]=27+x$, so $[A G C]=20$, which implies that $[C A D]=20+x$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6cb05c45fe7c141f3316g-5.jpg?height=666&width=700&top_left_y=163&top_left_x=688)

Now triangles $G A D$ and $G D B$ have the same altitude (from $G$ to $A B$ ), so their bases are proportional to their respective areas. In other words,

$$
\frac{A D}{D B}=\frac{[G A D]}{[G D B]}=\frac{x}{20-x}
$$

But $A D$ and $D B$ are also the bases of triangles $C A D$ and $C B D$, which have the same altitude (from $C$ to $A B)$. Hence

$$
\frac{A D}{D B}=\frac{[C A D]}{[C D B]}=\frac{20+x}{34+x}
$$

Equating these two fractions leads to the quadratic equation $34 x+x^{2}=400-x^{2}$; the only positive solution is $x=8$.