In a textbook the following problem example:
Two circles touch at point $M$. Through the point $M$ straight lines are drawn that intersect one circle at points $A_{1}$ and $B_{1}$, and the other at points $A_{2}$ and $B_{2}$. Prove that $A_{1}B_{1} \parallel A_{2}B_{2}$.
Has the next proof(its beginning):
Let's draw a tangent CD to one of the circles through the point M. This line will also be tangent to the second circle (prove this statement yourself)...
The figure itself:
After some attempts of "proving this statement myself" I went searching for the solution, and the only similar question I've found was this post with no answers.
You can show that $M$ is on the segment which ends are centers of both circle, if you have that, the end result is direct :) You can say that there is only one perpendicular line passing through a specific point.
Let us denote $O_1,r_1$ respectively the center and radius of the first circle, and same with the second. To show it we only need to show that $M = \frac{r_2}{r_1+r_2} O_1 + \frac{r_1}{r_1+r_2} O_2 $ or in affine geometry terms which should be more suitable for $8^{th}$ grade studs $$\overrightarrow{O_1M} + \overrightarrow{MO_2} = \overrightarrow{O_1 O_2}.$$
We have $|O_i M| = r_i$.
From the triangle inequality we have $$|O_1O_2| \leqslant r_1 + r_2.$$
We set $N = O_1 + \frac{r_1}{r_1 +r_2}\overrightarrow{O_1O_2}$, $$|O_1N| = \frac{r_1}{r_1 +r_2}|O_1O_2| \leqslant r_1$$ so $N \in C(O_1,r_1)$.
We also have $$|O_2 N| = (1- \frac{r_1}{r_1 +r_2})|O_1O_2| = \frac{r_2}{r_1 +r_2}|O_1O_2| \leqslant r_2$$ so $N \in C(O_2,r_2)$, hence $N$ lies in the intersection of the two disks, thus $N = M$, which proves the point :)