The triangle $ABC$ is acute, so the points $I$ and $O$ lie on the same side of the line $ED$. The condition that the points $D$, $E$, $I$ and $O$ lie on the same circle therefore implies that $\angle DOE = \angle DIE$.

Let us denote the angles of the triangle by $\alpha$, $\beta$ and $\gamma$ and let us express the angles $\angle DOE$ and $\angle DIE$ in these terms.

We have $\angle DOE = \angle EOC + \angle COD$. Since the central angle is always twice the inscribed angle, we get $\angle COE = 2\angle CBE$. The line $EB$ is the bisector of the angle $CBA$, so $\angle CBE = \frac{\beta}{2}$. Hence, $\angle COE = 2\angle CBE = \beta$.

Similarly, we have $\angle DOC = 2\angle DAC = 2 \cdot \frac{\alpha}{2} = \alpha$ and so $\angle DOE = \alpha + \beta$.

Also, $\angle DIE = \angle AIB = \pi - \angle BAI - \angle IBA = \pi - \frac{\alpha}{2} - \frac{\beta}{2}$.

The identity $\angle DOE = \angle DIE$ implies that $\pi = \frac{3}{2}(\alpha + \beta)$ or $\alpha + \beta = \frac{2\pi}{3}$.

We conclude that $\gamma = \pi - \alpha - \beta = \frac{\pi}{3}$.