The area of triangle $ABC$ is $2\sqrt{3}$, side $BC$ is $1$, and $\angle BCA = 60^\circ$. Point $D$ on side $AB$ is $3$ units away from point $B$, and $M$ is the intersection of $CD$ with the median $BE$. Find the ratio $BM : ME$.

#
From the equality $S_{A B C}=1 / 2 A C \cdot B C \sin \angle C$, we find that $A C=8 \sqrt{3}$. By the cosine theorem,

$$
A B^{2}=A C^{2}+B C^{2}-A C \cdot B C \cdot \sqrt{3}=192+1-24=169
$$

therefore, $A D=A B-B D=13-3=10$.

Draw a line through vertex $B$ parallel to $A C$. Let this line intersect the extension of segment $C D$ at point $K$. Triangle $B D K$ is similar to triangle $A D C$ with a similarity ratio of 0.3, so $B K=0.3 A C=0.6 C E$. Triangle $B M K$ is similar to triangle $E M C$, hence, $B M: M E=B K: C E=3: 5$.

## Answer

## $3: 5$.