[ Law of Sines [Sum of angles in a triangle. Theorem of the exterior angle.]

In triangle $ABC$, point $D$ lies on side $BC$, line $AD$ intersects the bisector of angle $C$ at point $O$. It is known that points $C, D$, and $O$ lie on a circle, the center of which is on side $AC$, $AC: AB = 3: 2$, and angle $DAC$ is three times angle $DAB$. Find the cosine of angle $ACB$.

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Let $\angle A C O=\gamma, \angle B A D=\alpha$. Express $\alpha$ in terms of $\gamma$ and apply the Law of Sines to triangle $A B C$.

Let $\angle C=2 \gamma, \angle B A D=\alpha, \angle C A D=3 \alpha, C E-$ be the diameter of the circumcircle $\omega$ of triangle $C D O$. Then $\angle O D E=\angle O C E=\gamma, \angle C D E=90^{\circ}$,

$\angle D E C=90^{\circ}-2 \gamma$. Point $A$ lies on the extension of segment $D O$ beyond point $O$, so it is farther from the center of $\omega$ than point $O$. Therefore, $\angle D E C$ is the exterior angle of triangle $A D E$, from which $\angle D E C=90^{\circ}-2 \gamma=3 \alpha+\gamma$, so $\alpha=30^{\circ}-\gamma$. Therefore, $\angle B=180^{\circ}-2 \gamma-4 \alpha=60^{\circ}+2 \gamma$.

By the Law of Sines and the problem's condition, $\frac{\sin 2 \gamma}{\sin \left(60^{\circ}+2 \gamma\right)}=\frac{2}{3}$. After obvious transformations, we get: $3 \sin 2 \gamma=\sqrt{3} \cos 2 \gamma+\sin 2 \gamma, \operatorname{tg} 2 \gamma=\frac{\sqrt{3}}{2}$, from which $\cos ^{2} 2 \gamma=\frac{1}{1+\operatorname{tg}^{2} 2 \gamma}=\frac{4}{7}$, and since $2 \gamma<90^{\circ}$ (as an acute angle of the right triangle $C D E$), $\cos 2 \gamma=\frac{2}{\sqrt{7}}$.

## Answer

$\frac{2}{\sqrt{7}}$.