A regular hexagon $ABCDEF$ has area $36$. Find the area of the region which lies in the overlap of the triangles $ACE$ and $BDF$.

$\text{(A) }3\qquad\text{(B) }9\qquad\text{(C) }12\qquad\text{(D) }18\qquad\text{(E) }24$
1. **Understanding the Problem:**
   We are given a regular hexagon \(ABCDEF\) with an area of 36. We need to find the area of the region that lies in the overlap of the triangles \(ACE\) and \(BDF\).

2. **Area of Triangles \(ACE\) and \(BDF\):**
   Since \(ACE\) and \(BDF\) are both equilateral triangles formed by connecting every other vertex of the hexagon, each of these triangles will have an area that is half of the area of the hexagon. 
   \[
   \text{Area of } \triangle ACE = \text{Area of } \triangle BDF = \frac{1}{2} \times 36 = 18
   \]

3. **Intersection of Triangles \(ACE\) and \(BDF\):**
   The intersection of these two triangles forms a smaller regular hexagon at the center of the original hexagon. This smaller hexagon is formed by the intersection of the medians of the original hexagon.

4. **Area of the Overlapping Region:**
   The smaller hexagon formed by the intersection of \(ACE\) and \(BDF\) is similar to the original hexagon and its side length is half of the side length of the original hexagon. The area of a hexagon is proportional to the square of its side length. Therefore, the area of the smaller hexagon is:
   \[
   \left(\frac{1}{2}\right)^2 \times 36 = \frac{1}{4} \times 36 = 9
   \]

5. **Conclusion:**
   The area of the region which lies in the overlap of the triangles \(ACE\) and \(BDF\) is 9.

The final answer is \(\boxed{9}\)