1. Identify the key points and lines:
- Let $BD$ be the median of triangle $ABC$, so $D$ is the midpoint of $AC$.
- Let $M$ be the midpoint of $BD$.
- Let line $\ell$ pass through $A$ and $M$, intersecting $BC$ at $E$.

2. **Determine the area of triangle $BDC$:**
- Since $BD$ is a median, it divides triangle $ABC$ into two triangles of equal area.
- Therefore, the area of triangle $BDC$ is $\frac{1}{2}S$.

3. **Determine the area of triangle $BME$:**
- Since $M$ is the midpoint of $BD$, $BM = MD$.
- The line $\ell$ through $A$ and $M$ divides triangle $BDC$ into two smaller triangles, $BME$ and $MEC$.
- By the sine area formula, the ratio of the areas of triangles $BME$ and $BDC$ is $\frac{1}{6}$.
- Therefore, the area of triangle $BME$ is:
$$[BME] = \frac{1}{6} \times [BDC] = \frac{1}{6} \times \frac{1}{2}S = \frac{1}{12}S$$

4. **Determine the area of quadrilateral $DMEC$:**
- The area of quadrilateral $DMEC$ is the remaining part of triangle $BDC$ after removing triangle $BME$.
- Therefore, the area of quadrilateral $DMEC$ is:
$$[DMEC] = [BDC] - [BME] = \frac{1}{2}S - \frac{1}{12}S = \frac{5}{12}S$$

5. **Determine the area of triangle $ADM$:**
- Since $M$ is the midpoint of $BD$, $AM$ is a median of triangle $ABD$.
- The median divides triangle $ABD$ into two triangles of equal area.
- Therefore, the area of triangle $ADM$ is:
$$[ADM] = \frac{1}{2} \times [ABD] = \frac{1}{2} \times \frac{1}{2}S = \frac{1}{4}S$$

6. **Determine the area of triangle $AMB$:**
- Similarly, the area of triangle $AMB$ is:
$$[AMB] = \frac{1}{2} \times [ABD] = \frac{1}{2} \times \frac{1}{2}S = \frac{1}{4}S$$