In the parallelogram $ABCD$, a line through $C$ intersects the diagonal $BD$ at $E$ and $AB$ at $F$. If $F$ is the midpoint of $AB$ and the area of $\vartriangle BEC$ is $100$, find the area of the quadrilateral $AFED$.
1. **Identify the given information and relationships:**
   - \(ABCD\) is a parallelogram.
   - \(F\) is the midpoint of \(AB\).
   - A line through \(C\) intersects diagonal \(BD\) at \(E\) and \(AB\) at \(F\).
   - The area of \(\triangle BEC\) is \(100\).

2. **Determine the relationship between triangles \(BEF\) and \(DEC\):**
   - Since \(F\) is the midpoint of \(AB\), \(AF = FB\).
   - Triangles \(BEF\) and \(DEC\) are similar because they share angle \(E\) and both have corresponding angles equal due to the parallel sides of the parallelogram.
   - The ratio of similarity is \(\frac{1}{2}\) because \(F\) is the midpoint of \(AB\).

3. **Calculate the area of \(\triangle DEC\):**
   - Since \(\triangle BEC\) and \(\triangle DEC\) are similar with a ratio of \(\frac{1}{2}\), the area of \(\triangle DEC\) is:
     \[
     [BEC] = \frac{1}{2}[DEC] \implies 100 = \frac{1}{2}[DEC] \implies [DEC] = 200
     \]

4. **Calculate the area of \(\triangle BEF\):**
   - The area of \(\triangle BEF\) is:
     \[
     [BEF] = \left(\frac{1}{2}\right)^2 [DEC] = \frac{1}{4} \times 200 = 50
     \]

5. **Calculate the area of parallelogram \(ABCD\):**
   - The area of parallelogram \(ABCD\) is twice the area of \(\triangle BDC\):
     \[
     [ABCD] = 2 \times [BDC]
     \]
   - Since \(\triangle BDC\) includes \(\triangle BEC\) and \(\triangle DEC\):
     \[
     [BDC] = [BEC] + [DEC] = 100 + 200 = 300
     \]
   - Therefore:
     \[
     [ABCD] = 2 \times 300 = 600
     \]

6. **Calculate the area of quadrilateral \(AFED\):**
   - The area of quadrilateral \(AFED\) is the area of parallelogram \(ABCD\) minus the areas of \(\triangle BEF\), \(\triangle BEC\), and \(\triangle DEC\):
     \[
     [AFED] = [ABCD] - [BEF] - [BEC] - [DEC] = 600 - 50 - 100 - 200 = 250
     \]

The final answer is \(\boxed{250}\)