3. Let $O$ be the center of the circumscribed circle $\mathcal{K}$ of an acute triangle $ABC$. The internal angle bisector at $A$ intersects the circle $\mathcal{K}$ again at point $D$, and the internal angle bisector at $B$ intersects the circle $\mathcal{K}$ again at point $E$. Let $I$ be the center of the inscribed circle of triangle $ABC$. What is the measure of angle $\angle A C B$ if points $D, E, O$, and $I$ lie on the same circle?
II/3. Since triangle $ABC$ is acute, points $I$ and $O$ lie on the same side of line $ED$. From the condition that points $D, E, I$, and $O$ lie on the same circle, it follows that $\angle DOE = \angle DIE$. Let's denote the angles in the triangle by $\alpha, \beta$, and $\gamma$ and express the angles $\angle DOE$ and $\angle DIE$ in terms of these.

We have $\angle DOE = \angle EOC + \angle COD$. Since a central angle is twice the inscribed angle, $\angle COE = 2 \angle CBE$. Line $EB$ is the angle bisector of $\angle CBA$, so $\angle CBE = \frac{\angle CBA}{2} = \frac{\beta}{2}$. Therefore, $\angle COE = 2 \angle CBE = \beta$. Similarly, $\angle DOC = 2 \angle DAC = 2 \cdot \frac{\alpha}{2} = \alpha$, and thus $\angle DOE = \alpha + \beta$.

We also have $\angle DIE = \angle AIB = \pi - \angle BAI - \angle IBA =$
![](https://cdn.mathpix.com/cropped/2024_06_07_f9d3b39f46a346efcf30g-20.jpg?height=620&width=665&top_left_y=978&top_left_x=1164)
$\pi - \frac{\alpha}{2} - \frac{\beta}{2}$. From the equality $\angle DOE = \angle DIE$, it follows that $\pi = \frac{3}{2}(\alpha + \beta)$ or $\alpha + \beta = \frac{2\pi}{3}$. Therefore, $\gamma = \pi - \alpha - \beta = \frac{\pi}{3}$.