We need to prove that $r_1 r_2 q = r q_1 q_2$. To do this, we will show that

$$\frac{r}{q} = \frac{r_1}{q_1} \cdot \frac{r_2}{q_2}$$

Let $s, s_1, s_2$ be the semiperimeters, and $\Delta, \Delta_1, \Delta_2$ the areas of the triangles $\triangle ABC, \triangle AMC, \triangle BMC$ respectively. Let $h$ be the common altitude from $C$ to $AB$.

The circle pairs $(r, q), (r_1, q_1), (r_2, q_2)$ are centrally similar with the common external homothety center $C$, the intersection of their common external tangents. Their pairwise homothety coefficients are equal to the distances of their tangency points from the vertex $C$:

$$\frac{r}{q} = \frac{s - AB}{s}, \quad \frac{r_1}{q_1} = \frac{s_1 - AM}{s_1}, \quad \frac{r_2}{q_2} = \frac{s_2 - BM}{s_2}$$

We need to show that:

$$\frac{s - AB}{s} = \frac{s_1 - AM}{s_1} \cdot \frac{s_2 - BM}{s_2}$$

Rewriting the above equation, we get:

$$1 - \frac{AB}{s} = \left(1 - \frac{AM}{s_1}\right) \left(1 - \frac{BM}{s_2}\right)$$

Expanding the right-hand side, we have:

$$1 - \frac{AB}{s} = 1 - \frac{AM}{s_1} - \frac{BM}{s_2} + \frac{AM}{s_1} \cdot \frac{BM}{s_2}$$

This simplifies to:

$$\frac{AB}{s} = \frac{AM}{s_1} + \frac{BM}{s_2} - \frac{AM}{s_1} \cdot \frac{BM}{s_2}$$

Substituting $AB = \frac{2 \Delta}{h}, AM = \frac{2 \Delta_1}{h}, BM = \frac{2 \Delta_2}{h}$ and $s = \frac{\Delta}{r}, s_1 = \frac{\Delta_1}{r_1}, s_2 = \frac{\Delta_2}{r_2}$, we get:

$$\frac{2r}{h} = \frac{2r_1}{h} + \frac{2r_2}{h} - \frac{4r_1 r_2}{h^2}$$

Reducing by $\frac{2h}{r_1 r_2}$ and rearranging, we obtain:

$$\frac{2}{h} = \frac{1}{r_1} + \frac{1}{r_2} - \frac{r}{r_1 r_2}$$

This relation has been proved in the referenced problems and confirms the required equality.

$\blacksquare$