7. In triangle $A B C$, side $B C$ is equal to 4, and angle $A C B$ is equal to $\frac{2 \pi}{3}$. Circle $\Gamma$ with radius $2 \sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that segment $M K$ is parallel to $A C$. Find the lengths of segments $C K, M K, A B$ and the area of triangle $C M N$.
Answer: $C K=2, M K=6, A B=4 \sqrt{13}, S_{\triangle C M N}=\frac{72 \sqrt{3}}{13}$.

Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ as $H$, and the angle $A B C$ as $\beta$.

The center of the circle inscribed in an angle lies on the angle bisector, so $\angle O C K = \frac{1}{2} \angle A C B = 60^{\circ}$. Since the radius to the point of tangency is perpendicular to the tangent, $C K = O K \operatorname{ctg} \angle O C K = 2 \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 2$. Therefore, the segment $B K$ is also equal to 2. By the sum of the angles in the quadrilateral $O L C K$, we find that $\angle L O K = 60^{\circ}$.

Since $O L \perp A C$ and $M K \| A C$, then $O L \perp M K$. Triangle $M O K$ is isosceles, and the height in it is also the angle bisector, so $\angle M O K = 2 \angle L O K = 120^{\circ}$. Hence, $M K = 2 \cdot M O \cdot \sin 60^{\circ} = 6$. Due to the parallelism of the lines $M K$ and $A C$, we get that $\angle M K B = 120^{\circ}$. Consider triangle $M K B$. By the cosine rule, $B M^{2} = B K^{2} + K M^{2} - 2 \cdot B K \cdot K M \cdot \cos 120^{\circ} = 52$, $B M = 2 \sqrt{13}$. By the sine rule, $\frac{M K}{\sin \beta} = \frac{B M}{\sin 120^{\circ}}$, from which $\sin \beta = \frac{3 \sqrt{3}}{2 \sqrt{13}} ; C H = B C \sin \beta = \frac{6 \sqrt{3}}{\sqrt{13}}$.

By the tangent-secant theorem, $B K^{2} = B M \cdot B N$, from which $B N = \frac{2}{\sqrt{13}} ; M N = B M - B N = \frac{24}{\sqrt{13}}$. Therefore, the area of triangle $C M N$ is $\frac{1}{2} C H \cdot M N = \frac{72 \sqrt{3}}{13}$.

Since $K M \| A C$, triangles $B K M$ and $B C A$ are similar, with the similarity ratio being $B K : B C = 1 : 2$. Hence, $\frac{B M}{B A} = \frac{1}{2}, A B = 4 \sqrt{13}$.