The problem involves the triangle $ABC$ with medians $AD$ and $BE$. The intersection of the medians is point $G$, and $F$ is a point on $AG$ such that $FG = \frac{1}{3} AG$. We are asked to find the value of the ratio $\frac{AF}{AD}$.

Solution:

1. In any triangle, the medians intersect at the centroid $G$, which divides each median into two segments in the ratio $2:1$, with the longer segment closer to the vertex.

2. Here, $F$ is on $AG$, and it is given that $FG = \frac{1}{3} AG$.

   Since $FG$ is $\frac{1}{3}$ of $AG$, the other segment $AF$ must be $\frac{2}{3}$ of $AG$.

3. The length of the entire median $AD$ is $AG + GD$, and since $G$ divides $AD$ in the ratio $2:1$, we have: $AG = \frac{2}{3} AD \quad \text{and} \quad GD = \frac{1}{3} AD.$

4. Since $AF = \frac{2}{3} AG$, substituting $AG$ in terms of $AD$, we get: $AF = \frac{2}{3} \times \frac{2}{3} AD = \frac{4}{9} AD.$

5. Therefore, the ratio $\frac{AF}{AD}$ is: $\frac{AF}{AD} = \frac{4}{9}.$

Thus, the correct answer is:

c) $\frac{4}{9}$.

Would you like more details or have any questions?

Here are some related questions:

1. How do you calculate the centroid of a triangle?
2. What are the properties of medians in a triangle?
3. How does the centroid divide each median of a triangle?
4. How can we use ratios to solve geometric problems involving medians?
5. What other important points of concurrency exist in a triangle?

Tip: In any triangle, the centroid divides each median in a consistent $2:1$ ratio, which is a key property for solving many problems involving medians.