In the drawing to the side, triangle $ABC$ is equilateral and $BD = CE = AF = \frac{AB}{3}$. The ratio $\frac{EG}{GD}$ can be written in the form $\frac{m}{n}, \operatorname{gcd}(m, n)=1$. What is the value of $m+n$?

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-37.jpg?height=460&width=516&top_left_y=1150&top_left_x=884)

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Solution

Notice that $\frac{E G}{G D}=\frac{[E B G]}{[B G D]}$. Now, $\frac{[E B G]}{[B C F]}=\frac{\frac{1}{2} E B \cdot B G \cdot \sin(\angle E B G)}{\frac{1}{2} B C \cdot B F \cdot \sin(\angle C B F)}$. Since $\angle E B G=\angle C B F$, we have

$$
\frac{[E B G]}{[B C F]}=\frac{E B \cdot B G}{B C \cdot B F}
$$

Similarly,

$$
\frac{[B G D]}{[A B F]}=\frac{B G \cdot B D}{B A \cdot B F}
$$

Dividing (1) by (2), we get $\frac{[E B G]}{[B G D]} \cdot \frac{[A B F]}{[B C F]}=\frac{E B \cdot B A}{B C \cdot B D}$. Finally, since $\frac{[A B F]}{[B C F]}=$ $\frac{A F}{C F}=\frac{1}{2}$, we have

$$
\frac{[E B G]}{[B G D]}=\frac{E B \cdot B A \cdot C F}{B C \cdot B D \cdot A F}=\frac{2}{3} \cdot 3 \cdot 2=4
$$

Therefore, $\frac{E G}{G D}=\frac{4}{1}$ and, hence, $m+n=5$.