Two congruent equilateral triangles $\triangle ABC$ and $\triangle DEF$ lie on the same side of line $BC$ so that $B$, $C$, $E$, and $F$ are collinear as shown. A line intersects $\overline{AB}$, $\overline{AC}$, $\overline{DE}$, and $\overline{EF}$ at $W$, $X$, $Y$, and $Z$, respectively, such that $\tfrac{AW}{BW} = \tfrac29$ , $\tfrac{AX}{CX} = \tfrac56$ , and $\tfrac{DY}{EY} = \tfrac92$. The ratio $\tfrac{EZ}{FZ}$ can then be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
[asy]
size(200);
defaultpen(linewidth(0.6));
real r = 3/11, s = 0.52, l = 33, d=5.5;
pair A = (l/2,l*sqrt(3)/2), B = origin, C = (l,0), D = (3*l/2+d,l*sqrt(3)/2), E = (l+d,0), F = (2*l+d,0);
pair W = r*B+(1-r)*A, X = s*C+(1-s)*A, Y = extension(W,X,D,E), Z = extension(W,X,E,F);
draw(E--D--F--B--A--C^^W--Z);
dot("$A$",A,N);
dot("$B$",B,S);
dot("$C$",C,S);
dot("$D$",D,N);
dot("$E$",E,S);
dot("$F$",F,S);
dot("$W$",W,0.6*NW);
dot("$X$",X,0.8*NE);
dot("$Y$",Y,dir(100));
dot("$Z$",Z,dir(70));
[/asy]
1. **Identify the given ratios and apply Menelaus' theorem:**
   - Given ratios:
     \[
     \frac{AW}{WB} = \frac{2}{9}, \quad \frac{AX}{CX} = \frac{5}{6}, \quad \frac{DY}{EY} = \frac{9}{2}
     \]
   - We need to find the ratio \(\frac{EZ}{FZ}\).

2. **Apply Menelaus' theorem to \(\triangle ABZ\) with transversal \(WXZ\):**
   - Menelaus' theorem states that for a triangle \( \triangle ABC \) and a transversal intersecting \( \overline{AB}, \overline{BC}, \overline{CA} \) at points \( D, E, F \) respectively, the following holds:
     \[
     \frac{AD}{DB} \cdot \frac{BE}{EC} \cdot \frac{CF}{FA} = 1
     \]
   - For \(\triangle ABZ\) with transversal \(WXZ\):
     \[
     \frac{AW}{WB} \cdot \frac{BZ}{ZC} \cdot \frac{CX}{XA} = 1
     \]
   - Substitute the given ratios:
     \[
     \frac{2}{9} \cdot \frac{BZ}{ZC} \cdot \frac{6}{5} = 1
     \]
   - Solve for \(\frac{BZ}{ZC}\):
     \[
     \frac{2}{9} \cdot \frac{6}{5} \cdot \frac{BZ}{ZC} = 1 \implies \frac{12}{45} \cdot \frac{BZ}{ZC} = 1 \implies \frac{BZ}{ZC} = \frac{45}{12} = \frac{15}{4}
     \]

3. **Apply Menelaus' theorem to \(\triangle ABE\) with transversal \(WVZ\):**
   - For \(\triangle ABE\) with transversal \(WVZ\):
     \[
     \frac{AW}{WB} \cdot \frac{BZ}{ZE} \cdot \frac{EV}{VA} = 1
     \]
   - Since \(AV = VE\) (as \(W\) and \(Y\) are symmetric with respect to the midline of parallelogram \(ABED\)):
     \[
     \frac{AW}{WB} \cdot \frac{BZ}{ZE} \cdot 1 = 1
     \]
   - Substitute the given ratio:
     \[
     \frac{2}{9} \cdot \frac{BZ}{ZE} = 1 \implies \frac{BZ}{ZE} = \frac{9}{2}
     \]

4. **Determine the ratio \(\frac{EZ}{FZ}\):**
   - Let \(BZ = 45\). Then:
     \[
     ZC = \frac{BZ}{\frac{15}{4}} = \frac{45}{\frac{15}{4}} = 12
     \]
   - Since \(BZ = 45\) and \(ZC = 12\), \(BC = 33\).
   - \(EZ = 10\) (since \(BZ = 45\) and \(\frac{BZ}{ZE} = \frac{9}{2}\)).
   - \(EF = 33\) (since \(BC = 33\)).
   - \(FZ = EF - EZ = 33 - 10 = 23\).
   - Therefore:
     \[
     \frac{EZ}{FZ} = \frac{10}{23}
     \]

The final answer is \(\boxed{33}\).