By $E_3, D_3$ we denote points that are symmetrical to $E_1, D_1$ with respect to AB and AC, respectively (Fig. 45). It follows from symmetry that $AE_1 = AE_2 = AE_3$, and CA is a bisector of an angle that is created by the lines $CE_3$ and $CE_2$. Then we have
$$\angle(CE_3, E_3A) = \angle(E_1E_3, E_3A) = \angle(AE_1, E_1E_3) = \angle(AE_1, E_1C) = \angle(CE_2, E_2A).$$

Thus points A, $E_2$, $E_3$ and C are cyclic.
In the same way points A, $D_2$, $D_3$ and B are cyclic. Then the circumcircle of the triangle $\triangle ABD_2$ coincides with the circumcircle of $\triangle AD_3D_2$. In the same way, the circumcircle of $\triangle ACE_2$ coincides with the circumcircle of $\triangle AE_3E_2$. Since $AE_3$ and $AD_2$ are symmetrical to the same line with respect to AB, then points A, $E_3$, $D_2$ are collinear. In the same way, A, $E_2$, $D_3$ are collinear. Moreover,
$$k = \frac{AD_2}{AE_3} = \frac{AD_1}{AE_1} = \frac{AD_3}{AE_2}.$$
Thus $\triangle AE_2E_3 \sim \triangle AD_3D_2$, and their circumcircles are tangent to each other, because under homothety $H_A^k$ one of the circles maps to the other circle and they have a common point A.

Pic. 45