## Task B-2.5.

In triangle $A B C$, point $D$ is on side $\overline{A B}$ such that $|A D|:|D B|=3: 4$. On segment $\overline{C D}$, point $E$ is such that $|C E|:|E D|=3: 4$. A line parallel to $A E$ passes through point $D$ and intersects $\overline{B C}$ at point $G$. Determine the ratio $|C G|:|G B|$.
## Solution.

Let's draw the diagram.

![](https://cdn.mathpix.com/cropped/2024_05_30_a25c851e6461b5bf7f1cg-08.jpg?height=660&width=1011&top_left_y=1229&top_left_x=451)

Let the line $A E$ intersect the side $\overline{B C}$ at $F$ and let $DG$ be parallel to $EF$. By Thales' theorem, we have

$$
\frac{|C F|}{|F G|}=\frac{3}{4}=\frac{9}{12}
$$

It also holds that: $\frac{|F G|}{|G B|}=\frac{|A D|}{|D B|}=\frac{3}{4}=\frac{12}{16}$. Therefore,

$$
\begin{aligned}
& |C F|:|F G|:|G B|=9: 12: 16 \\
& |C F|=9 k,|F G|=12 k,|G B|=16 k \\
\Rightarrow & \frac{|C G|}{|G B|}=\frac{|C F|+|F G|}{|G B|}=\frac{9 k+12 k}{16 k}=\frac{21}{16}
\end{aligned}
$$

$$
\begin{aligned}
& 1 \text { point } \\
& 1 \text { point }
\end{aligned}
$$

1 point