Let $ABCD$ be a rhombus of sides $AB = BC = CD= DA = 13$. On the side $AB$ construct the rhombus  $BAFC$ outside $ABCD$ and such that the side $AF$  is parallel to the diagonal $BD$ of $ABCD$. If the area of $BAFE$ is equal to $65$, calculate the area of $ABCD$.
1. **Identify the properties of the rhombus \(ABCD\)**:
   - All sides are equal: \(AB = BC = CD = DA = 13\).
   - Diagonals bisect each other at right angles.
   - Let \(H\) be the intersection of the diagonals \(AC\) and \(BD\).

2. **Construct the rhombus \(BAFC\) outside \(ABCD\)**:
   - Given that \(AF\) is parallel to the diagonal \(BD\) of \(ABCD\).
   - The area of \(BAFC\) is given as \(65\).

3. **Determine the length of \(FG\)**:
   - Since \(BAFC\) is a rhombus with side length \(13\) and area \(65\), the height \(FG\) from \(F\) to \(BE\) can be calculated as:
     \[
     \text{Area} = \text{side} \times \text{height} \implies 65 = 13 \times FG \implies FG = \frac{65}{13} = 5
     \]

4. **Relate \(FG\) to the diagonals of \(ABCD\)**:
   - Since \(AF\) is parallel to \(BD\) and \(FG\) is perpendicular to \(BE\), \(FG\) represents half the length of the diagonal \(AC\) of \(ABCD\).
   - Therefore, \(AC = 2 \times FG = 2 \times 5 = 10\).

5. **Use the Pythagorean theorem in \(\triangle ABH\)**:
   - In \(\triangle ABH\), \(AB = 13\) and \(AH = \frac{AC}{2} = 5\).
   - Calculate \(BH\):
     \[
     BH = \sqrt{AB^2 - AH^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12
     \]

6. **Determine the length of diagonal \(BD\)**:
   - Since \(H\) is the midpoint of \(BD\), \(BD = 2 \times BH = 2 \times 12 = 24\).

7. **Calculate the area of rhombus \(ABCD\)**:
   - The area of a rhombus is given by \(\frac{1}{2} \times \text{product of diagonals}\).
   - Here, the diagonals are \(AC = 10\) and \(BD = 24\):
     \[
     \text{Area of } ABCD = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 10 \times 24 = 120
     \]

The final answer is \(\boxed{120}\).