The diagram below shows the regular hexagon $BCEGHJ$ surrounded by the rectangle $ADFI$. Let $\theta$ be the measure of the acute angle between the side $\overline{EG}$ of the hexagon and the diagonal of the rectangle $\overline{AF}$. There are relatively prime positive integers $m$ and $n$ so that $\sin^2\theta  = \tfrac{m}{n}$. Find $m + n$.

[asy]
import graph; size(3.2cm); 
real labelscalefactor = 0.5; 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
draw((-1,3)--(-1,2)--(-0.13,1.5)--(0.73,2)--(0.73,3)--(-0.13,3.5)--cycle); 
draw((-1,3)--(-1,2)); 
draw((-1,2)--(-0.13,1.5)); 
draw((-0.13,1.5)--(0.73,2)); 
draw((0.73,2)--(0.73,3)); 
draw((0.73,3)--(-0.13,3.5)); 
draw((-0.13,3.5)--(-1,3)); 
draw((-1,3.5)--(0.73,3.5)); 
draw((0.73,3.5)--(0.73,1.5)); 
draw((-1,1.5)--(0.73,1.5)); 
draw((-1,3.5)--(-1,1.5)); 
label("$ A $",(-1.4,3.9),SE*labelscalefactor); 
label("$ B $",(-1.4,3.28),SE*labelscalefactor); 
label("$ C $",(-1.4,2.29),SE*labelscalefactor); 
label("$ D $",(-1.4,1.45),SE*labelscalefactor); 
label("$ E $",(-0.3,1.4),SE*labelscalefactor); 
label("$ F $",(0.8,1.45),SE*labelscalefactor); 
label("$ G $",(0.8,2.24),SE*labelscalefactor); 
label("$ H $",(0.8,3.26),SE*labelscalefactor); 
label("$ I $",(0.8,3.9),SE*labelscalefactor); 
label("$ J $",(-0.25,3.9),SE*labelscalefactor); [/asy]
1. **Assume the side length of the regular hexagon \(BCEGHJ\) is 1.**
   - Since it is a regular hexagon, each internal angle is \(120^\circ\).
   - The distance between opposite vertices (e.g., \(B\) and \(E\)) is twice the side length, i.e., \(2\).

2. **Determine the coordinates of the vertices of the hexagon.**
   - Place the hexagon in the coordinate plane with the center at the origin \(O\).
   - The vertices can be calculated using trigonometric functions:
     \[
     B = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \quad C = \left(-1, 0\right), \quad E = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right), \quad G = \left(1, 0\right), \quad H = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \quad J = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)
     \]

3. **Calculate the coordinates of the rectangle \(ADFI\).**
   - The rectangle surrounds the hexagon, so its sides are parallel to the coordinate axes.
   - The vertices of the rectangle are:
     \[
     A = \left(-1, \frac{\sqrt{3}}{2}\right), \quad D = \left(-1, -\frac{\sqrt{3}}{2}\right), \quad F = \left(1, -\frac{\sqrt{3}}{2}\right), \quad I = \left(1, \frac{\sqrt{3}}{2}\right)
     \]

4. **Find the equation of the diagonal \(AF\).**
   - The slope of \(AF\) is:
     \[
     \text{slope of } AF = \frac{-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}}{1 - (-1)} = \frac{-\sqrt{3}}{2}
     \]
   - The equation of the line \(AF\) is:
     \[
     y = -\frac{\sqrt{3}}{2}x + \frac{\sqrt{3}}{2}
     \]

5. **Find the angle \(\theta\) between \(EG\) and \(AF\).**
   - The slope of \(EG\) is:
     \[
     \text{slope of } EG = \frac{0 - (-\frac{\sqrt{3}}{2})}{1 - \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}
     \]
   - The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
     \[
     \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
     \]
     where \(m_1 = \sqrt{3}\) and \(m_2 = -\frac{\sqrt{3}}{2}\):
     \[
     \tan \theta = \left| \frac{\sqrt{3} - \left(-\frac{\sqrt{3}}{2}\right)}{1 + \sqrt{3} \cdot \left(-\frac{\sqrt{3}}{2}\right)} \right| = \left| \frac{\sqrt{3} + \frac{\sqrt{3}}{2}}{1 - \frac{3}{2}} \right| = \left| \frac{\frac{3\sqrt{3}}{2}}{-\frac{1}{2}} \right| = \left| -3\sqrt{3} \right| = 3\sqrt{3}
     \]
   - Since \(\tan \theta = 3\sqrt{3}\), we use the identity:
     \[
     \sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{(3\sqrt{3})^2}{1 + (3\sqrt{3})^2} = \frac{27}{1 + 27} = \frac{27}{28}
     \]

6. **Identify \(m\) and \(n\) and find \(m + n\).**
   - Here, \(m = 27\) and \(n = 28\).
   - Therefore, \(m + n = 27 + 28 = 55\).

The final answer is \(\boxed{55}\)