In the diagram below, points $D, E$, and $F$ are on the inside of equilateral $\vartriangle ABC$ such that $D$ is on $\overline{AE}, E$ is on $\overline{CF}, F$ is on $\overline{BD}$, and the triangles $\vartriangle AEC, \vartriangle BDA$, and $\vartriangle CFB$ are congruent. Given that $AB = 10$ and $DE = 6$, the perimeter of $\vartriangle BDA$ is $\frac{a+b\sqrt{c}}{d}$, where $a, b, c$, and $d$ are positive integers, $b$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a + b + c + d$.
[img]https://cdn.artofproblemsolving.com/attachments/8/6/98da82fc1c26fa13883a47ba6d45a015622b20.png[/img]
1. **Identify the given information and the goal:**
   - We have an equilateral triangle \( \triangle ABC \) with side length \( AB = 10 \).
   - Points \( D, E, \) and \( F \) are inside the triangle such that \( D \) is on \( \overline{AE} \), \( E \) is on \( \overline{CF} \), and \( F \) is on \( \overline{BD} \).
   - The triangles \( \triangle AEC \), \( \triangle BDA \), and \( \triangle CFB \) are congruent.
   - We need to find the perimeter of \( \triangle BDA \) in the form \( \frac{a + b\sqrt{c}}{d} \) and then find \( a + b + c + d \).

2. **Analyze the congruent triangles:**
   - Since \( \triangle AEC \), \( \triangle BDA \), and \( \triangle CFB \) are congruent, they share the same angles and side lengths.
   - Given \( \angle DAC + \angle ECA = 60^\circ \) and \( \angle ECA = \angle DAB \), it follows that \( \angle AEC = 120^\circ \).

3. **Determine the properties of \( \triangle DEF \):**
   - Since \( \angle AEC = 120^\circ \), \( \angle DEF = 60^\circ \), making \( \triangle DEF \) equilateral.
   - Given \( DE = 6 \), all sides of \( \triangle DEF \) are 6.

4. **Use the Law of Cosines in \( \triangle ABD \):**
   - Let \( AD = x \) and \( BF = x \) (since \( \triangle BDA \) and \( \triangle CFB \) are congruent).
   - Apply the Law of Cosines in \( \triangle ABD \):
     \[
     AB^2 = AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos(120^\circ)
     \]
     \[
     10^2 = x^2 + (6 + x)^2 - 2 \cdot x \cdot (6 + x) \cdot \left(-\frac{1}{2}\right)
     \]
     \[
     100 = x^2 + (6 + x)^2 + x(6 + x)
     \]
     \[
     100 = x^2 + 36 + 12x + x^2 + 6x + x^2
     \]
     \[
     100 = 3x^2 + 18x + 36
     \]
     \[
     3x^2 + 18x + 36 = 100
     \]
     \[
     3x^2 + 18x - 64 = 0
     \]

5. **Solve the quadratic equation:**
   - Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 3 \cdot (-64)}}{2 \cdot 3}
     \]
     \[
     x = \frac{-18 \pm \sqrt{324 + 768}}{6}
     \]
     \[
     x = \frac{-18 \pm \sqrt{1092}}{6}
     \]
     \[
     x = \frac{-18 \pm 6\sqrt{273}}{6}
     \]
     \[
     x = -3 \pm \sqrt{273}
     \]
     - Since \( x \) must be positive, we take \( x = -3 + \sqrt{273} \).

6. **Calculate the perimeter of \( \triangle BDA \):**
   - The sides of \( \triangle BDA \) are \( AD = x \), \( BD = 6 + x \), and \( AB = 10 \).
   - Perimeter:
     \[
     \text{Perimeter} = x + (6 + x) + 10 = 2x + 16
     \]
     \[
     \text{Perimeter} = 2(-3 + \sqrt{273}) + 16
     \]
     \[
     \text{Perimeter} = -6 + 2\sqrt{273} + 16
     \]
     \[
     \text{Perimeter} = 10 + 2\sqrt{273}
     \]
     - Expressed in the form \( \frac{a + b\sqrt{c}}{d} \):
       \[
       \frac{30 + 2\sqrt{273}}{3}
       \]

7. **Identify \( a, b, c, \) and \( d \):**
   - \( a = 30 \), \( b = 2 \), \( c = 273 \), \( d = 3 \).

8. **Calculate \( a + b + c + d \):**
   \[
   a + b + c + d = 30 + 2 + 273 + 3 = 308
   \]

The final answer is \( \boxed{308} \)