4. Given a parallelogram $A B C D$. A line passing through point $D$ intersects segment $A C$ at point $G$, segment $B C$ at point $F$, and ray $A B$ at point $E$ such that triangles $BEF$ and $CGF$ have the same area. Determine the ratio $|A G|:|G C|$.
SOLUTION. From Fig. 1, it can be seen that triangles $A G D$ and $C G F$ are similar according to the theorem

![](https://cdn.mathpix.com/cropped/2024_04_17_00132d3cc3e29ab78bbdg-2.jpg?height=442&width=623&top_left_y=510&top_left_x=708)

uu. The similarity ratio $k$ is equal to the sought ratio $|A G|:|G C|$. If we denote $b=|A D|, x=|D G|$, and $y=|C G|$, then $|G F|=x / k$ and $|C F|=b / k$, from which

$$
|F B|=|B C|-|F B|=b-\frac{b}{k}=(k-1) \frac{b}{k}
$$

and

$$
|D F|=|D G|+|G F|=x+\frac{x}{k}=(k+1) \frac{x}{k} .
$$

From the similarity of triangles $B E F \sim C D F$, we get

$$
|E F|=\frac{|D F| \cdot|B F|}{|C F|}=\frac{k^{2}-1}{k} \cdot x .
$$

From the equality of the areas of triangles $B E F$ and $C G F$, it follows that

$$
|F B| \cdot|F E|=|F C| \cdot|F G|,
$$

which, after substitution, yields

$$
\frac{k-1}{k} \cdot b \cdot \frac{k^{2}-1}{k} \cdot x=\frac{b}{k} \cdot \frac{x}{k} .
$$

Thus, $k^{3}-k^{2}-k+1=1$, and since $k \neq 0$, we get the quadratic equation $k^{2}-k-1=0$ for the sought $k$. The problem is satisfied by its positive root $k=\frac{1}{2}(1+\sqrt{5})$.