$[\quad$ Law of Cosines $\quad]$ [ Similar Auxiliary Triangles ]

On the side $AB$ of triangle $ABC$, a point $D$ is chosen such that $CD=\sqrt{13}$ and $\sin \angle ACD : \sin \angle BCD = 4 : 3$. Through the midpoint of segment $CD$, a line is drawn intersecting sides $AC$ and $BC$ at points $M$ and $N$ respectively. It is known that $\angle C = 120^\circ$, the area of triangle $MCN$ is $3\sqrt{3}$, and the distance from point $M$ to line $AB$ is twice the distance from point $N$ to the same line. Find the area of triangle $ABC$.
Solving the trigonometric equation, find the angles $A C D$ and $B C D$, then use the cosine theorem to prove that the segment $C D$ passes through the midpoint $M N$.

## Solution

Let $C M=x, C N=y, \angle A C D=\alpha, \angle B C D=\beta$. According to the condition $3 \sin \alpha=4 \sin \left(120^{\circ}-\alpha\right)$, from which $\operatorname{tg} \alpha=2 \sqrt{3}, \quad \cos \alpha=\frac{1}{\sqrt{13}}, \quad \sin \alpha=\frac{2 \sqrt{3}}{\sqrt{13}}, \quad \sin \beta=\frac{3}{4} \sin \alpha=\frac{3 \sqrt{3}}{2 \sqrt{13}}$. Let $P$ be the midpoint of segment $C D$. Since $S_{M C P}+S_{N C P}=S_{M C N}=3 \sqrt{3}$, then $C M \cdot C P \sin \alpha+C N \cdot C P \sin \beta=$ $C M \cdot C N \sin 120^{\circ}=6 \sqrt{3}$, that is, $x \cdot \frac{\sqrt{13}}{2} \cdot \frac{2 \sqrt{3}}{\sqrt{13}}+y \cdot \frac{\sqrt{13}}{2} \cdot \frac{3 \sqrt{3}}{2 \sqrt{13}}=x y \cdot \frac{\sqrt{3}}{2}=6 \sqrt{3}$, or $4 x+3 y=24$, $x y=12$, from which $x=3, y=4$. By the cosine theorem

$M N^{2}=C M^{2}+C N^{2}-2 C M \cdot C N \cos 120^{\circ}=37, M P^{2}=C M^{2}+C P^{2}-2 C M \cdot C P \cos \alpha=9+{ }^{13} / 4-3={ }^{37} / 4$.

Therefore, $P$ is the midpoint of segment $C D$. The diagonals of quadrilateral $M C N D$ are bisected by their intersection point, so $M C N D$ is a parallelogram. Thus, $D N \| A M$ and $D M \| B N$, meaning triangles $B N D$ and $D M A$ are similar, and since the heights from vertices $N$ and $M$ are in the ratio $1: 2$, the similarity ratio of these triangles is $1 / 2$. Therefore, $B N=1 / 2 D M=1 / 2 C N=2$ and $A M=2 D N=2 C M=6$, from which $A C=9$ and $B C=6$. Therefore, $S_{A B C}=1 / 2 A C \cdot B C \sin 120^{\circ}=\frac{27 \sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-03.jpg?height=314&width=623&top_left_y=186&top_left_x=723)

## Answer

$\frac{27 \sqrt{3}}{2}$.