We need following fact.

Lemma: Given a point $D$ on side $BC$ of the triangle $ABC$. Let $\omega_1(O_1, r_1), \omega_2(O_2, r_2)$ be incircles of triangles $ACD$, $ADB$ and $\rho_1(O_3, r_3), \rho_2(O_4, r_4)$ be circles inscribed in these triangles externally, which tangent at an internal point of $BC$. Then lines $O_1O_2$, $O_3O_4$, $BC$ pass through a point.

Proof of lemma: Denote $BC \cap O_3O_4 = X$ and let's prove that $X \in O_1O_2$. From the well known property that a line passing centers of circles inscribed in an angle is bisector of the angle, it follows $A = (O_1O_3) \cap (O_2O_4)$, $D = (O_1O_4) \cap (O_2O_3)$. Since the line $O_3O_2$ intersects sides of the triangle $AO_1O_4$,

$$\text{by Menelaus theorem we get } 1 = \frac{O_3O_1}{O_3A} \cdot \frac{DO_4}{DO_1} \cdot \frac{O_2A}{O_2O_4} = \frac{O_3A - O_1A}{O_3A} \cdot \frac{DO_4}{DO_1} \cdot \frac{O_2A}{O_4A - O_2A}$$

$$= \left(1 - \frac{O_1A}{O_3A}\right) \cdot \frac{DO_4}{DO_1} \cdot \left(\frac{1}{\frac{O_4A}{O_2A} - 1}\right) = \left(1 - \frac{r_1}{r_3}\right) \left(\frac{r_4}{r_1}\right) \left(\frac{1}{\frac{r_4}{r_2} - 1}\right)$$

$$= \frac{(r_3 - r_1)}{r_3} \cdot \frac{r_4}{r_1} \cdot \frac{r_2}{r_4 - r_2} = \frac{(r_3 - r_1) r_4 r_2}{r_3 r_1 (r_4 - r_2)} \quad (*)$$

To prove $X \in (O_1O_2)$ is equivalent to proving by Menelaus theorem:

$$\frac{XO_3}{XO_4} \cdot \frac{O_1A}{O_1O_3} \cdot \frac{O_2O_4}{O_2A} = 1.$$

Furthermore,

$$\frac{XO_3}{XO_4} = \frac{r_3}{r_4}, \quad \frac{O_1A}{O_1O_3} = \frac{O_1A}{O_3A - AO_1} = \frac{1}{\frac{O_3A}{O_1A} - 1} = \frac{1}{\frac{r_3}{r_1} - 1} = \frac{r_1}{r_3 - r_1},$$

$$\frac{O_2O_4}{O_2A} = \frac{AO_4 - AO_2}{O_2A} = \frac{AO_4}{AO_2} - 1 = \frac{r_4}{r_2} - 1 = \frac{r_4 - r_2}{r_2}$$

and by $(*)$,

$$\frac{XO_3}{XO_4} \cdot \frac{O_1A}{O_1O_3} \cdot \frac{O_2O_4}{O_2A} = \frac{r_3}{r_4} \cdot \frac{r_1}{r_3 - r_1} \cdot \frac{r_4 - r_2}{r_2} = 1.$$

This proves that $X \in O_1O_2$ and points $O_1O_2$, $O_3O_4$, $BC$ pass through a point.

Now solve the problem. Let $l$ be common tangent to circles $\omega_1, \omega_2, \omega_3, \omega_4$ which is different from $BC$. Denote $l \cap BC = X$. Centers of circles $\omega_1, \omega_2, \omega_3, \omega_4$ lie on the bisector of angle formed by $l$, $BC$ and the point $X$ lies on this bisector too.

Using the lemma for triangles $ABE$, $ADF$, $AEC$ we conclude that centers of circles $\rho_1, \rho_2, \rho_3, \rho_4$ and the point $X$ lie on a line. Reflecting $BC$ about this line we get a line which is common tangent to circles $\rho_1, \rho_2, \rho_3, \rho_4$. Thus we have the desired result.