Let us write $\angle ACB = \gamma$. Then $\angle CBA = 2\gamma$ and $\angle BAC = \pi - 3\gamma$. If we draw a figure where $\angle CBA = 2\angle ACB$, we notice that $IA_1C_1$ is a right triangle. Let us prove this.
Since $ABA_1B_1$ is a cyclic quadrilateral we have $\angle C_1A_1I = \angle B_1A_1A = \angle B_1BA = \frac{\angle BCA}{2} = \gamma$. The angle $\angle A_1IC$ is equal to $\pi - \angle CIA$. Since $\angle CIA = \pi - \angle CAI - \angle ACI$, we get
$$\angle A_1IC = \angle CAI + \angle ACI = \frac{\pi - 3\gamma}{2} + \frac{\gamma}{2} = \frac{\pi}{2} - \gamma.$$

So, $\angle A_1IC + \angle C_1A_1I = \frac{\pi}{2}$ and $\angle IC_1A_1 = \frac{\pi}{2}$. Thales' theorem states that the circumcentre of the triangle $IA_1C_1$ (i.e. the point S) is the midpoint of the segment $IA_1$. The central angle is twice the inscribed angle, so $\angle C_1SI = 2\angle C_1A_1I = 2\frac{\angle CBA}{2} = 2\gamma$. The quadrilateral $ABA_1C_1$ is cyclic, so $\angle CA_1A = \angle CBA = 2\gamma$. Hence, $CA_1$ is parallel to DS. But D is the midpoint of the segment CE and S is the midpoint of the segment $IA_1$, so DS is the midsegment of the quadrilateral $SIA_1C_1$. Since DS is parallel to $CA_1$ it must also be parallel to EI. So, $\angle EIA = \angle DSA = 2\gamma$.

For the inner angles of the triangle ABI we have $\angle BAI = \angle BAC = \frac{\pi-3\gamma}{2}$, $\angle IBA = \angle CBA = \frac{\pi-3\gamma}{2}$. Since the sum of the inner angles in an arbitrary triangle is equal to $\pi$ we get $\frac{\pi-3\gamma}{2} + \gamma + \pi - 2\gamma = \pi$ or $\gamma = \frac{\pi}{5}$.

The angles of the triangle ABC measure $\angle BAC = \angle CBA = \frac{2\pi}{5}$ and $\angle ACB = \frac{\pi}{5}$.