In the notations of the posted picture we have, also considering signs:
$$
\begin{aligned}
-\frac{A'B}{A'C} &=
\frac{BA'}{A'C} =
\frac{BK}{AC}=
\frac{BC}{AC} &(\Delta A'BK &\sim\Delta A'CA)\ ,\\[5mm]
-\frac{B'C}{B'A} &=
\frac{CB'}{B'A} =
\frac{BC}{AF} =
\frac{BC}{AC} &(\Delta B'CB &\sim\Delta B'AF)\ ,\\[5mm]
-\frac{MA}{MB} &=
\frac{AM}{MB} =
\frac{AM\cdot AB}{MB\cdot AB} =
\frac{AC^2}{BC^2}
\ .
\\[3mm]
&\qquad\text{ All together:}
\\[3mm]
\frac{A'B}{A'C}\cdot
\frac{B'C}{B'A}\cdot
\frac{MA}{MB}
&=-\frac{BC}{AC}\cdot \frac{BC}{AC}\cdot \frac{AC^2}{BC^2} =-1\ .
\end{aligned}
$$
Using Ceva's theorem, the lines $AA'$, $BB'$, $CM$ are concurrent.
$\square$
We are done with the given problem in the given very special situation, but for the sake of the geometric taste, let us place the result in a more general context.
Generalization:
Proposition: Let $ABC$ be a triangle, on the sides are constructed in the exterior of the triangle squares. Among the many vertices we need only two, $F,K$, the new vertices of the two squares on $CA$, respectively $CB$, so that $FA\perp CA$ and $KB\perp CB$. Let $M$ be the projection of $C$ on $AB$, so that $CM$ is a height of $\Delta ABC$.
Then the lines $AK$, $BF$, $CM$ are concurrent.
Before we start the proof, recall the following theorem of Pappus.
It deals with two lines, on the one line we have the points, in order, $A,B,C$, on the other line corresponding points $a,b,c$, also in the written order. Now we build intersections $X,Y,Z$ (see the linked wiki page), and the theorem insures that they are collinear.
In projective geometry it is also allowed, to use points at infinity. Such a point is given by specifying a (classical) line $l$, and then its point at infinity, denoted here by $\infty_l$, is "the limit point" (in a suitable model) when we move a point on $l$ in some chosen coordinate system on the line to infinity. Incidence properties are important only in projective geometry. An other line $k$ also goes through $\infty_l$ iff $k$ and $l$ are parallel / have same direction.
Proof of the Proposition:
With the above in mind, we can apply Pappus for our configuration, below in picture a general case:
Here, $f$ is the line $AF$, $k$ is the line $AK$, and we apply the theorem for:
- the points $A,F,\infty_f$ on the line $f$,
- the points $B,K,\infty_k$ on the line $k$.
Doing so we build the three intersection points:
- $BF\cap AK$, this is the green point.
- $B\infty_f\cap A\infty_k=H$, which is the orthocenter of the triangle $\Delta ABC$, since $B\infty_f\|f$ is perpendicular on the side $AC$, and $A\infty_k\| k$ is perpendicular on the side $BC$.
- $K\infty_f\cap A\infty_k=H$,
and it remains to show that $TH\perp AB$, because then this line passing through the orthocenter $H$ is the same as $CM$. For this, let us denote by $a,b,c$ the sides of $\Delta ABC$, by $A,B,C$ also abusively the angles in the same vertices, and by $h_a,h_b,h_c$ the heights (each seen as line, and as length, depending on the context).
We compute the slope of $TH$, in the system with axes parallel to the lines $AC$ and $BC$. The projection of $KB$ on $AC$ is after a $90^\circ$-rotation the same as the one of $BC$ (same length as $AC$) on the perpendicular direction, so it is $h_b$. Its projection on $BC$ is $h_a$.
The segment $CH$ has proportional projections, $CH\cdot \cos\widehat {HCA}=CH\sin A$ on $AC$, and $CH\sin B$ on $BC$. We check this proportionality in one line, using:
$$
\frac{CH\sin A}{CH\sin B}
=
\frac{\sin A}{\sin B}
=
\frac ab
=
\frac{h_b}{h_a}
\ .
$$
(Last equality uses the formula for the area, so $ah_a/2=bh_b/2$.)
This shows that $CH$ and $TH$ have same direction, the direction perpendicular to $AB$. This concludes the proof of the Proposition.
$\square$
Note: Applying Pappus in the same manner on the initial configuration is much easier, almost a two lines proof.
Best Answer
Let $r_1$, $r_2$, and $r_3$ be the radii of $o_1$, $o_2$, and $o_3$, respectively. Let $i\in\{1,2,3\}$. The indices are considered modulo $3$. Let $S_{i-1}S_{i+1}$ be an internal common tangent of $o_{i-1}$ and $o_{i+1}$ with $S_{i-1}\in o_{i-1}$ and $S_{i+1}\in o_{i+1}$. Prove that $O_{i-1}S_{i-1}T_i$ and $O_{i+1}S_{i+1}T_i$ are similar triangles. This shows that $$\frac{O_{i-1}A_i}{A_iO_{i+1}}=\frac{r_{i-1}}{r_{i+1}}\,.$$