5. Circle G with radius $4 \sqrt{3}$ 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 $A C, C L=4, B K=3$. Find $\angle A C B$, the lengths of segments $M K, A B$ and the area of triangle $B K N$.
Answer: $\angle A C B=\frac{2 \pi}{3}, M K=12, A B=7 \sqrt{21}, S_{\triangle B K N}=\frac{3 \sqrt{3}}{7}$.

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

Since the radius drawn to the point of tangency is perpendicular to the tangent, $\operatorname{to} \operatorname{tg} \angle O C K=$ $\frac{O K}{C K}=\sqrt{3}$ and $\angle O C K=60^{\circ}$. The center of the circle inscribed in an angle lies on the bisector of that angle, so $\angle A C B=2 \angle O C K=120^{\circ}$. Then, by the sum of the angles of 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, the height in it is also the bisector, hence, $\angle M O K=2 \angle L O K=120^{\circ}$. From this, $M K=2 \cdot M O \cdot \sin 60^{\circ}=$ 12.

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 theorem, $B M^{2}=B K^{2}+K M^{2}-2 \cdot B K \cdot K M \cdot \cos 120^{\circ}=189$, $B M=3 \sqrt{21}$. By the sine theorem, $\frac{M K}{\sin \beta}=\frac{B M}{\sin 120^{\circ}}$, from which $\sin \beta=\frac{2}{\sqrt{7}}$.

By the tangent and secant theorem, $B K^{2}=B M \cdot B N$, from which $B N=\frac{3}{\sqrt{21}}$. Then the area of triangle $B K N$ is $\frac{1}{2} \cdot B K \cdot B N \cdot \sin \beta=\frac{3 \sqrt{3}}{7}$.

Since $K M \| A C$, triangles $B K M$ and $B C A$ are similar, with the similarity coefficient equal to $B K: B C=3: 7$. From this, it follows that $\frac{B M}{B A}=\frac{3}{7}, \frac{3 \sqrt{21}}{B A}=\frac{3}{7}, A B=7 \sqrt{21}$.