IRL ABooklet

Connect $A$ to $D$ and $G$. Because $\triangle ABC$ and $\triangle BCD$ are equilateral, $\angle BAD = \angle BDA = 30^\circ$ and $|AD| = \sqrt{3}$. Similarly, $\angle GAF = 30^\circ$ and $|AG| = \sqrt{3}$. We now have $\angle BDG = \angle BDA + \angle ADG = 30^\circ + \angle ADG$

and $\angle BAF = \angle BAD + \angle DAG + \angle GAF = 60^\circ + \angle DAG$. We will show that $\angle BDG > \angle BAF$ by proving $\cos(\angle ADG) < \cos(30^\circ + \angle DAG)$.
Letting $H$ be the midpoint of $DG$ and using triangle $ADH$ we find
$$\cos(\angle ADG) = \frac{|DH|}{|DA|} = \frac{1}{2\sqrt{3}} = \frac{2\sqrt{3}}{12}.$$

Because $\cos(30^\circ + \angle DAG) = \cos(30^\circ)\cos(\angle DAG) - \sin(30^\circ)\sin(\angle DAG)$, we next wish to find $\cos(\angle DAG)$ and $\sin(\angle DAG)$. The Cosine Rule for

$\triangle ADG$ yields $\cos(\angle DAG) = 5/6$. Because $\angle DAG < 180^\circ$, this implies $\sin(\angle DAG) = \sqrt{11}/6$. Hence
$$\cos(30^\circ + \angle DAG) = \frac{\sqrt{3}}{2} \cdot \frac{5}{6} - \frac{1}{2} \cdot \frac{\sqrt{11}}{6} = \frac{5\sqrt{3} - \sqrt{11}}{12}.$$
To prove $2\sqrt{3} < 5\sqrt{3} - \sqrt{11}$ we rewrite this as $\sqrt{11} < 3\sqrt{3}$ and then square. This shows that $\cos(\angle ADG) < \cos(30^\circ + \angle DAG)$, which means that $\angle ADG > 30^\circ + \angle DAG$, and so $\angle BDG > \angle BAF$.