1. **Problem statement:** In triangle $\triangle ABC$, $D$ is the midpoint of side $BC$. Point $E$ lies on side $AC$ such that $CE = \frac{AC}{3}$. Lines $BE$ and $AD$ intersect at point $G$. We need to find the ratio $\frac{AG}{GD}$. 2. **Key concepts:** - Since $D$ is the midpoint of $BC$, $BD = DC = \frac{BC}{2}$. - Point $E$ divides $AC$ so that $CE = \frac{1}{3} AC$, meaning $AE = \frac{2}{3} AC$. - $G$ is the intersection of $BE$ and $AD$, so $G$ lies on both segments. 3. **Approach:** Use coordinate geometry or vector method to find the ratio $\frac{AG}{GD}$. Assign coordinates: - Let $A = (0,0)$. - Let $C = (c,0)$ on the x-axis. - Since $D$ is midpoint of $BC$, and $B$ is unknown, assign $B = (x_b,y_b)$. 4. **Coordinates of points:** - $D$ midpoint of $BC$: $D = \left(\frac{x_b + c}{2}, \frac{y_b + 0}{2}\right) = \left(\frac{x_b + c}{2}, \frac{y_b}{2}\right)$. - $E$ lies on $AC$ such that $CE = \frac{1}{3} AC$, so $E$ divides $AC$ in ratio $AE:EC = 2:1$. - Since $A=(0,0)$ and $C=(c,0)$, $E = \left(\frac{2}{3}c, 0\right)$. 5. **Parametric equations:** - Line $AD$: from $A$ to $D$ is $\vec{r}_{AD}(t) = tD = \left(t \frac{x_b + c}{2}, t \frac{y_b}{2}\right)$ for $t \in [0,1]$. - Line $BE$: from $B$ to $E$ is $\vec{r}_{BE}(s) = B + s(E - B) = \left(x_b + s\left(\frac{2}{3}c - x_b\right), y_b + s(0 - y_b)\right)$ for $s \in [0,1]$. 6. **Find intersection $G$:** Set $\vec{r}_{AD}(t) = \vec{r}_{BE}(s)$: $$ \begin{cases} t \frac{x_b + c}{2} = x_b + s\left(\frac{2}{3}c - x_b\right) \\ t \frac{y_b}{2} = y_b + s(-y_b) \end{cases} $$ 7. **From second equation:** $$ t \frac{y_b}{2} = y_b (1 - s) \implies t = 2 \frac{1 - s}{1} = 2(1 - s) $$ 8. **From first equation:** $$ t \frac{x_b + c}{2} = x_b + s\left(\frac{2}{3}c - x_b\right) $$ Substitute $t = 2(1 - s)$: $$ 2(1 - s) \frac{x_b + c}{2} = x_b + s\left(\frac{2}{3}c - x_b\right) \\ (1 - s)(x_b + c) = x_b + s\left(\frac{2}{3}c - x_b\right) $$ 9. **Expand and simplify:** $$ (x_b + c) - s(x_b + c) = x_b + s\left(\frac{2}{3}c - x_b\right) \\ (x_b + c) - s(x_b + c) - x_b = s\left(\frac{2}{3}c - x_b\right) \\ c - s(x_b + c) = s\left(\frac{2}{3}c - x_b\right) \\ c = s\left(\frac{2}{3}c - x_b + x_b + c\right) = s\left(\frac{2}{3}c + c\right) = s \frac{5}{3} c $$ 10. **Solve for $s$:** $$ s = \frac{c}{\frac{5}{3} c} = \frac{c}{\frac{5}{3} c} = \frac{3}{5} $$ 11. **Find $t$:** $$ t = 2(1 - s) = 2\left(1 - \frac{3}{5}\right) = 2 \times \frac{2}{5} = \frac{4}{5} $$ 12. **Interpretation:** $t$ is the parameter along $AD$ from $A$ to $D$. Since $t = \frac{4}{5}$, point $G$ divides $AD$ in ratio $AG : GD = 4 : 1$. **Final answer:** $$\boxed{\frac{AG}{GD} = 4 : 1}$$