7. The inscribed circle of triangle $A B C$ with radius 3 touches side $A C$ at point $D$. On the extensions of sides $A C$ and $B C$ beyond point $C$, points $K$ and $L$ are taken respectively, such that angle $C K L$ is $30^{\circ}$. The lengths of segments $A K$ and $B L$ are equal to the semiperimeter of triangle $A B C$. On the circumcircle of triangle $A B C$, a point $M$ is chosen such that $C M \| K L$. Find the cosine of angle $A C B$, if the length of segment $D M$ is $6 . \quad$ (16 points)
Solution. We will prove that $\angle OMC = 90^{\circ}$. Construct a circle $S$ with diameter $OC$. Denote the intersection point (different from $C$) of this circle with the circumcircle of triangle $ABC$ as $M_1$. Denote the intersection point (different from $C$) of circle $S$ with the line parallel to $KL$ as $M_2$. We need to prove that $M_1 = M_2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-05.jpg?height=959&width=1219&top_left_y=100&top_left_x=156)

We will prove the similarity of triangles $ADM_1$ and $BNM_1$. According to the properties of inscribed angles, we have $\angle M_1AC = \angle M_1BC \Rightarrow \angle M_1AB = \angle M_1BN; \angle M_1NC = \angle M_1DC \Rightarrow \angle ADM_1 = \angle BNM_1$. Therefore, $\triangle ADM_1 \sim \triangle BNM_1$ by two angles. Then $\frac{DM_1}{M_1N} = \frac{AD}{BN}$. Since the lengths of segments $AK$ and $BL$ are equal to the semiperimeter of triangle $ABC$, we have $AD = CL, BN = CK$. Thus, $\frac{DM_1}{M_1N} = \frac{CL}{CK}$.

Let $\angle ACM_2 = \angle CKL = \alpha, \angle KCL = \beta$.

Then $\angle ACB = \beta, \angle M_2CB = \alpha + \beta$. For triangles $DM_2C$ and $M_2NC$, apply the Law of Sines: $\frac{DM_2}{\sin \alpha} = \frac{M_2C}{\sin \angle M_2DC}; \frac{NM_2}{\sin (\alpha + \beta)} = \frac{M_2C}{\sin \angle M_2NC}$. Since $\angle M_2DC = \angle M_2NC$, we have $\frac{DM_2}{NM_2} = \frac{\sin \alpha}{\sin (\alpha + \beta)}$. Since $\angle CLK = 180 - (\alpha + \beta)$, applying the Law of Sines to triangle $CLK$, we get $\frac{CL}{CK} = \frac{\sin \alpha}{\sin (\alpha + \beta)}$. Thus, $\frac{DM_2}{NM_2} = \frac{CL}{CK} = \frac{DM_1}{NM_1}$. Points $D, N, M_1, M_2$ lie on the same circle, so $M_1 = M_2$.

Triangle $OMC$ is a right triangle, the radius of the circumcircle is $CO / 2$. This same circle is circumscribed around triangle $CDM$.

$$
OC = \frac{OD}{\sin (\beta / 2)} = \frac{3}{\sin (\beta / 2)}.
$$

$\angle CKL$ equals $\angle DCM$. Then $\sin CKL = \frac{DM}{2R_{\text{op}}} = \frac{DM}{OC} = 0.5$. Therefore, $\sin (\beta / 2) = 0.25, \cos \beta = 1 - 2 \sin^2 (\beta / 2) = 7 / 8 = 0.875$.

## Answer: 0.875