Write $\angle ABD = \alpha$. Since $AO_1BD$ is a cyclic quadrilateral, we have $\angle AO_1D = \angle ABD = \alpha$. The central angle $\angle AO_1C$ in the circle $K_1$ is twice the size of the inscribed angle $\angle ABC$, so $\angle AO_1C = 2\alpha$.

$$\begin{aligned} \angle DO_1C &= \angle AO_1C - \angle AO_1D \\ &= 2\alpha - \alpha = \alpha. \end{aligned}$$
The segment $AB$ is perpendicular to the segment $O_1O_2$ and the quadrilateral $AO_1BO_2$ is a deltoid. So,
$$\begin{aligned} \angle ABO_1 &= \angle O_1AB = \frac{\pi}{2} - \angle O_2O_1A \\ &= \frac{\pi}{2} - \angle AO_1D - \angle DO_1C = \frac{\pi}{2} - 2\alpha. \end{aligned}$$

Since $AO_1BD$ is a cyclic quadrilateral, we have $\angle ADO_1 = \angle ABO_1 = \frac{\pi}{2} - 2\alpha$. Since $F$ is the midpoint of the segment $AE$, we have $\angle AFO_1 = \frac{\pi}{2}$. This implies $\angle FO_1D = \frac{\pi}{2} - \angle FDO_1 = 2\alpha$ and
$$\angle AO_1F = \angle DO_1F - \angle DO_1A = 2\alpha - \alpha = \alpha.$$
We have shown that $\angle AO_1F = \alpha = \angle AO_1C = \angle DO_1C$, so the lines $O_1A$ and $O_1D$ divide the angle $CO_1F$ into three equal parts.