[ Law of Sines ] [ Law of Cosines ] In triangle $A B C$, medians $A N$ and $C M$ are drawn, $\angle A B C=120^{\circ}$. The circle passing through points $A, M$ and $N$ also passes through point $C$. The radius of this circle is 7. Find the area of triangle $A B C$.
Prove that the given triangle is isosceles, find $A N$, and to find the sides of the given triangle, apply the Law of Cosines to triangle $A B N$.

## Solution

By the theorem of the midline of a triangle, $M N \| A C$, so $A M N C$ is a trapezoid. Since the trapezoid is inscribed in a circle, it is isosceles. Since $A B=2 \cdot A M=2 \cdot C N=B C$, the triangle $A B C$ is also isosceles, so $\angle B A C=\angle B C A=30^{\circ}$.

If $R$ is the radius of the circle, then

$$
A N=2 R \cdot \sin \angle A C N=2 R \cdot \sin 30^{\circ}=R=7
$$

Let $C N=A M=x$. Then $A B=B C=2 x$. By the Law of Cosines in triangle $A B N$, we have:

$$
A N^{2}=A B^{2}+B N^{2}-2 \cdot A B \cdot B N \cdot \cos 120^{\circ} \text{, or } 49=4 x^{2}+x^{2}+2 x^{2}
$$

From this, we find that $x^{2}=7$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot A B \cdot B C \cdot \sin \angle A B C=\frac{1}{2} \cdot 2 x \cdot 2 x \cdot \sin 120^{\circ}=2 x^{2} \cdot \frac{\sqrt{3}}{2}=x^{2} \sqrt{3}=7 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-36.jpg?height=343&width=878&top_left_y=2032&top_left_x=590)

## Answer

$7 \sqrt{3}$.