Inspiratio!

Quadrilateral EGFB is cyclic because angles B and G add to 180 degrees. This means BE=BF.

$\triangle BFH$ is congruent to the $\triangle BEJ$ . $\triangle BGJ$ is congruent to the $\triangle GHI$ . This means that the area of the quadrilateral EGFB is equal to the area of the equilateral $\triangle BGI$ . Segment $BG = \dfrac{ \sqrt{7} }{3} DC$ . Area of the rhombus is twice the area of the equilateral triangle with the same side length. The ratio of the area of the quadrilateral EGFB to area of the rhombus ABCD is $\dfrac{ 7}{2*9}=\dfrac{ 7}{18}$

Figure 1 If we use an isometric grid as shown on figure 1, we see that $\dfrac{EB}{AB} =\dfrac{BF}{BC}=\dfrac{7}{15}$ (see at the end of the solution for a more elaborate explanation)

Now, $\triangle GBE$ and $\triangle GBA$ share a common height from vertex $G$ , thus, $\begin{aligned} & \dfrac{\left[ GEB \right]}{\left[ GAB \right]}=\dfrac{EB}{AB}=\dfrac{7}{15} \\ & \Rightarrow \left[ GEB \right]=\dfrac{7}{15}\left[ GAB \right]=\dfrac{7}{15}\left[ DAB \right]=\dfrac{7}{15}\left( \dfrac{1}{2}\left[ ABCD \right] \right)=\dfrac{7}{30}\left[ ABCD \right] \ \ \ \ \ (1)\\ \end{aligned}$ Likewise, $\triangle GBF$ and $\triangle GBC$ share a common height from vertex G, thus, $\begin{aligned} & \dfrac{\left[ GBF \right]}{\left[ GBC \right]}=\dfrac{BF}{BC}=\dfrac{7}{15} \\ & \Rightarrow \left[ GBF \right]=\dfrac{7}{15}\left[ GBC \right] \ \ \ \ \ (2)\\ \end{aligned}$ Next, $\begin{aligned} & \dfrac{\left[ GBC \right]}{\left[ DBC \right]}=\dfrac{\cancel{\dfrac{1}{2}CB}\cdot CG\cdot \bcancel{\sin C}}{\cancel{\dfrac{1}{2}CB}\cdot CD\cdot \bcancel{\sin C}}=\dfrac{CG}{CD}=\dfrac{2}{3} \\ & \Rightarrow \left[ GBC \right]=\dfrac{2}{3}\left[ DBC \right] \ \ \ \ \ (3)\\ \end{aligned}$ $\left( 2 \right),\left( 3 \right)\Rightarrow \left[ GBF \right]=\dfrac{7}{15}\left( \dfrac{2}{3}\left[ DBC \right] \right)=\dfrac{14}{45}\left( \dfrac{1}{2}\left[ ABCD \right] \right)\Rightarrow \left[ GBF \right]=\dfrac{7}{45}\left[ ABCD \right] \ \ \ \ \ (4)$ Finally, $\left[ EGFB \right]=\left[ GEB \right]+\left[ GBF \right]\overset{\left( 1 \right),\left( 4 \right)}{\mathop{=}}\,\dfrac{7}{30}\left[ ABCD \right]+\dfrac{7}{45}\left[ ABCD \right]=\dfrac{7}{18}\left[ ABCD \right]$ i.e. $\dfrac{\left[ EGFB \right]}{\left[ ABCD \right]}=\dfrac{7}{18}$ For the answer, $a=7$ , $b=18$ , thus, $a+b=\boxed{25}$ .

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Details to justify that in the grid, $\angle EGF$ is indeed $60{}^\circ$ and $\angle BGF=30{}^\circ$ , i.e. $F$ and $E$ are the correct (given) points.

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Figure 2

Focusing around point $G$ , triangle congruency verifies that $\angle ΕGF=\alpha +\gamma=\beta +\gamma=60{}^\circ$ (figure 2).

Figure 3 Furthermore, since $\angle ΕBF+\angle EGF=120{}^\circ +60{}^\circ =180{}^\circ$ , $EGFB$ is a cyclic quadrilateral, thus $\angle BGF=\angle BEF=30{}^\circ$ (figure 3).

The shaded area can be cut and rearranged to fill out $7$ out of $18$ equilateral triangles as follows:

which makes the ratio $\cfrac{7}{18}$ , so $a = 7$ , $b = 18$ , and $a + b = \boxed{25}$ .