In triangle $\triangle A B C$, point $F$ is on side $A C$ and $F C=2 A F$. If $G$ is the midpoint of segment $B F$ and $E$ is the intersection point of the line passing through $A$ and $G$ with segment $B C$, calculate the ratio $\frac{E C}{E B}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-06.jpg?height=430&width=575&top_left_y=1189&top_left_x=691)
We have $\frac{F C}{A F}=2$. Now, draw the segment $F H$, parallel to the segment $A E$ where $H$ is on the segment $B C$, as shown in the following figure.

The triangles $\triangle A E C$ and $\triangle F H C$ are similar because they have parallel sides. This implies that $C H=2 E H$.

On the other hand, the triangles $\triangle B F H$ and $\triangle B G E$ are also similar, as they have parallel sides. From this similarity and the fact that $G$ is the midpoint of the segment $B F$, we conclude that $E$ is the midpoint of the segment $B H$.

Thus, $B E=E H$ and, therefore, $E C=E H+C H=E H+2 E H=3 E H=3 E B$. Consequently, $\frac{E C}{E B}=3$.

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-26.jpg?height=431&width=561&top_left_y=283&top_left_x=706)