IRL_ABooklet_2023

Connect $A$ to $D$ and $G$. Because $△ABC$ and $△BCD$ 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

$△ADG$ 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 $\angle BDG>\angle BAF$.