Triangle $ABC$ has sidelengths $AB=1$, $BC=\sqrt{3}$, and $AC=2$. Points $D,E$, and $F$ are chosen on $AB, BC$, and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$. Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a, b) = 1$, find $a + b$. (Here $[DEF]$ denotes the area of triangle $DEF$.) 

[i]Proposed by Vismay Sharan[/i]
1. **Identify the coordinates of points \( A, B, \) and \( C \):**
   - Place \( A \) at the origin, \( A = (0, 0) \).
   - Place \( B \) at \( (1, 0) \) since \( AB = 1 \).
   - Place \( C \) at \( (x, y) \) such that \( AC = 2 \) and \( BC = \sqrt{3} \).

2. **Determine the coordinates of \( C \):**
   - Using the distance formula for \( AC \):
     \[
     x^2 + y^2 = 4
     \]
   - Using the distance formula for \( BC \):
     \[
     (x-1)^2 + y^2 = 3
     \]
   - Solving these equations simultaneously:
     \[
     (x-1)^2 + y^2 = 3 \implies x^2 - 2x + 1 + y^2 = 3 \implies x^2 + y^2 - 2x + 1 = 3
     \]
     \[
     4 - 2x + 1 = 3 \implies 2x = 2 \implies x = 1
     \]
     \[
     y^2 = 4 - 1^2 = 3 \implies y = \sqrt{3}
     \]
   - Therefore, \( C = (1, \sqrt{3}) \).

3. **Determine the coordinates of points \( D, E, \) and \( F \):**
   - Let \( D \) be on \( AB \) at \( (x, 0) \).
   - Let \( E \) be on \( BC \) at \( (1, y) \).
   - Let \( F \) be on \( AC \) at \( (t, t\sqrt{3}) \).

4. **Use the given conditions \( \angle EDF = \angle DFA = 90^\circ \):**
   - The slopes of \( ED \) and \( DF \) must be negative reciprocals.
   - Slope of \( ED \):
     \[
     \text{slope of } ED = \frac{y - 0}{1 - x} = \frac{y}{1 - x}
     \]
   - Slope of \( DF \):
     \[
     \text{slope of } DF = \frac{t\sqrt{3} - 0}{t - x} = \frac{t\sqrt{3}}{t - x}
     \]
   - Since \( \angle EDF = 90^\circ \):
     \[
     \frac{y}{1 - x} \cdot \frac{t\sqrt{3}}{t - x} = -1
     \]

5. **Maximize the area of \( \triangle DEF \):**
   - The area of \( \triangle DEF \) is given by:
     \[
     [DEF] = \frac{1}{2} \left| x(y - t\sqrt{3}) + 1(t\sqrt{3} - 0) + t(0 - y) \right|
     \]
   - Simplify the expression:
     \[
     [DEF] = \frac{1}{2} \left| xy - xt\sqrt{3} + t\sqrt{3} - ty \right|
     \]
     \[
     [DEF] = \frac{1}{2} \left| t\sqrt{3}(1 - x) - y(t - x) \right|
     \]

6. **Find the maximum value of \( [DEF]^2 \):**
   - The maximum value of \( [DEF] \) occurs when \( x = \frac{1}{2} \):
     \[
     [DEF] = \frac{\sqrt{3}}{8}
     \]
   - Therefore, the maximum value of \( [DEF]^2 \) is:
     \[
     \left( \frac{\sqrt{3}}{8} \right)^2 = \frac{3}{64}
     \]

The final answer is \( \boxed{67} \)