The diagram below shows equilateral $\triangle ABC$ with side length $2$. Point $D$ lies on ray $\overrightarrow{BC}$ so that $CD = 4$. Points $E$ and $F$ lie on $\overline{AB}$ and $\overline{AC}$, respectively, so that $E$, $F$, and $D$ are collinear, and the area of $\triangle AEF$ is half of the area of $\triangle ABC$. Then $\tfrac{AE}{AF}=\tfrac m n$, where $m$ and $n$ are relatively prime positive integers. Find $m + 2n$.
[asy]
import math;
size(7cm);
pen dps = fontsize(10);
defaultpen(dps);
dotfactor=4;
pair A,B,C,D,E,F;
B=origin;
C=(2,0);
D=(6,0);
A=(1,sqrt(3));
E=(1/3,sqrt(3)/3);
F=extension(A,C,E,D);
draw(C--A--B--D,linewidth(1.1));
draw(E--D,linewidth(.7));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
label("$E$",E,NW);
label("$F$",F,NE);
[/asy]
1. **Assigning Mass Points:**
   - We start by assigning mass points to the vertices of $\triangle ABD$. 
   - Assign a mass of $4$ to point $B$ and a mass of $2$ to point $D$ so that the mass at $C$ (which is the sum of the masses at $B$ and $D$) is $6$.

2. **Finding the Mass at Point $A$:**
   - Let the mass at point $A$ be denoted by $A$.
   - Using the mass point system, the ratio $\frac{AE}{EB}$ is given by $\frac{4}{A}$.
   - Therefore, $\frac{AE}{AB} = \frac{4}{A+4}$, which implies $AE = \frac{8}{A+4}$.

3. **Finding $AF$:**
   - Similarly, for point $F$, $\frac{AF}{FC}$ is given by $\frac{2}{A}$.
   - Therefore, $\frac{AF}{AC} = \frac{2}{A+2}$, which implies $AF = \frac{4}{A+2}$.

4. **Using the Area Condition:**
   - The area of $\triangle AEF$ is half of the area of $\triangle ABC$.
   - The area of $\triangle ABC$ is $\frac{\sqrt{3}}{4} \times 2^2 = \sqrt{3}$.
   - Therefore, the area of $\triangle AEF$ is $\frac{\sqrt{3}}{2}$.
   - Using the formula for the area of a triangle, $\frac{1}{2} \times AE \times AF \times \sin 60^\circ = \frac{\sqrt{3}}{2}$.
   - Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we have:
     \[
     \frac{1}{2} \times AE \times AF \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}
     \]
     \[
     AE \times AF = 2
     \]

5. **Solving for $A$:**
   - Plugging in the expressions for $AE$ and $AF$:
     \[
     \frac{8}{A+4} \times \frac{4}{A+2} = 2
     \]
     \[
     \frac{32}{(A+4)(A+2)} = 2
     \]
     \[
     32 = 2(A^2 + 6A + 8)
     \]
     \[
     32 = 2A^2 + 12A + 16
     \]
     \[
     2A^2 + 12A + 16 = 32
     \]
     \[
     2A^2 + 12A - 16 = 0
     \]
     \[
     A^2 + 6A - 8 = 0
     \]
     \[
     A = \frac{-6 \pm \sqrt{36 + 32}}{2}
     \]
     \[
     A = \frac{-6 \pm \sqrt{68}}{2}
     \]
     \[
     A = \frac{-6 \pm 2\sqrt{17}}{2}
     \]
     \[
     A = -3 \pm \sqrt{17}
     \]
   - Since $A$ must be positive, we take $A = -3 + \sqrt{17}$.

6. **Finding $AE$ and $AF$:**
   - $AE = \frac{8}{A+4} = \frac{8}{-3 + \sqrt{17} + 4} = \frac{8}{1 + \sqrt{17}} = \frac{8(1 - \sqrt{17})}{(1 + \sqrt{17})(1 - \sqrt{17})} = \frac{8(1 - \sqrt{17})}{-16} = \frac{8 - 8\sqrt{17}}{-16} = \frac{1 - \sqrt{17}}{2}$
   - $AF = \frac{4}{A+2} = \frac{4}{-3 + \sqrt{17} + 2} = \frac{4}{-1 + \sqrt{17}} = \frac{4(-1 - \sqrt{17})}{(-1 + \sqrt{17})(-1 - \sqrt{17})} = \frac{4(-1 - \sqrt{17})}{-16} = \frac{4 - 4\sqrt{17}}{-16} = \frac{1 - \sqrt{17}}{4}$

7. **Finding the Ratio $\frac{AE}{AF}$:**
   - $\frac{AE}{AF} = \frac{\frac{1 - \sqrt{17}}{2}}{\frac{1 - \sqrt{17}}{4}} = \frac{4}{2} = 2$

8. **Finding $m + 2n$:**
   - Since $\frac{AE}{AF} = \frac{m}{n}$, we have $m = 2$ and $n = 1$.
   - Therefore, $m + 2n = 2 + 2 \times 1 = 4$.

The final answer is $\boxed{4}$