The area of the triangle $ABC$ shown in the figure is   $1$ unit. Points $D$ and $E$ lie on sides $AC$ and $BC$ respectively,   and also are its ''one third'' points closer to $C$. Let $F$ be that $AE$ and $G$ are the midpoints of segment $BD$. What is the area of the marked quadrilateral $ABGF$?
[img]https://cdn.artofproblemsolving.com/attachments/4/e/305673f429c86bbc58a8d40272dd6c9a8f0ab2.png[/img]
1. **Define the coordinates of the vertices:**
   Given $\triangle ABC$ with vertices $A(0,0)$, $B(3b,0)$, and $C(3a,3c)$.

2. **Determine the coordinates of points $D$ and $E$:**
   - Point $D$ is the "one third" point on $AC$ closer to $C$. Therefore, $D$ divides $AC$ in the ratio $1:2$.
     \[
     D = \left(\frac{2 \cdot 3a + 1 \cdot 0}{3}, \frac{2 \cdot 3c + 1 \cdot 0}{3}\right) = (2a, 2c)
     \]
   - Point $E$ is the "one third" point on $BC$ closer to $C$. Therefore, $E$ divides $BC$ in the ratio $1:2$.
     \[
     E = \left(\frac{2 \cdot 3a + 1 \cdot 3b}{3}, \frac{2 \cdot 3c + 1 \cdot 0}{3}\right) = (2a + b, 2c)
     \]

3. **Determine the coordinates of points $F$ and $G$:**
   - Point $F$ is the midpoint of $AE$.
     \[
     F = \left(\frac{0 + (2a + b)}{2}, \frac{0 + 2c}{2}\right) = \left(a + \frac{b}{2}, c\right)
     \]
   - Point $G$ is the midpoint of $BD$.
     \[
     G = \left(\frac{3b + 2a}{2}, \frac{0 + 2c}{2}\right) = \left(a + \frac{3b}{2}, c\right)
     \]

4. **Calculate the area of quadrilateral $ABGF$:**
   - The area of quadrilateral $ABGF$ can be found by subtracting the areas of triangles $ABF$ and $BGF$ from the area of $\triangle ABC$.
   - The area of $\triangle ABC$ is given as $1$ unit.

5. **Calculate the area of $\triangle ABF$:**
   - Using the coordinates $A(0,0)$, $B(3b,0)$, and $F(a + \frac{b}{2}, c)$, the area of $\triangle ABF$ is:
     \[
     \text{Area of } \triangle ABF = \frac{1}{2} \left| 0(0 - c) + 3b(c - 0) + (a + \frac{b}{2})(0 - 0) \right| = \frac{1}{2} \left| 3bc \right| = \frac{3bc}{2}
     \]

6. **Calculate the area of $\triangle BGF$:**
   - Using the coordinates $B(3b,0)$, $G(a + \frac{3b}{2}, c)$, and $F(a + \frac{b}{2}, c)$, the area of $\triangle BGF$ is:
     \[
     \text{Area of } \triangle BGF = \frac{1}{2} \left| 3b(c - c) + (a + \frac{3b}{2})(c - c) + (a + \frac{b}{2})(0 - 0) \right| = 0
     \]
     Since $G$ and $F$ lie on the same horizontal line, the area of $\triangle BGF$ is zero.

7. **Calculate the area of quadrilateral $ABGF$:**
   - The area of quadrilateral $ABGF$ is:
     \[
     \text{Area of } ABGF = \text{Area of } \triangle ABC - \text{Area of } \triangle ABF - \text{Area of } \triangle BGF
     \]
     \[
     \text{Area of } ABGF = 1 - \frac{3bc}{2} - 0 = 1 - \frac{3bc}{2}
     \]

Since the area of $\triangle ABC$ is given as $1$ unit, and the area of $\triangle ABF$ is $\frac{3bc}{2}$, we need to find the correct area of $ABGF$.

The final answer is $\boxed{\frac{4}{9}}$