Triangles with a common incircle I am given two triangles that intersect at 6 points, which share an incircle of radius $r$.
There are also 6 circles each tangent to two sides of one triangle from the inside, and tangent to a side of the other triangle from the outside. I will number these circles 1 through 6, $R_i$ is the radius of circle $i$.

I need to prove $R_1 R_3 R_5 = R_2 R_4 R_6$.
I know the inner circle with radius $r$ is an excircle of each of the 6 small triangles, so
$$r=\frac{R_i \cdot s_i}{s_i - a_i}$$ for all $i$ between 1 and 6, where $s_i$ is the semiperimeter of each small circle, and $a_i$ is the side of the triangle tangent to the circle with radius $r$.
How can I continue from here?
 A: Let us use the following notation of points:

We will use index notation modulo six, with representatives $1,2,3,4,5,6$ for the indices, and denote by


*

*$R_i$ the radius of the incircle of $\Delta A_iA_{i,i+1}A_{i-1,i}$,

*$A_{ij}$ also the angle in $A_{ij}$ in $\Delta A_iA_{i,i+1}A_{i-1,i}$, for each of the vertices labeled with two indices,


and we observe that the notations are compatible. For instance, the angles in $A_{12}$ in the two triangles $\Delta A_{12}A_{61}A_1$ and
$\Delta A_{12}A_{23}A_2$ are equal.
The proof and the generalization are now immediate, given the following lemma:


Lemma: Let $\Delta ABC$ be a triangle, let $r$ be the radius of the incircle, and let $r_a$ be the radius of the ex-circle tangent to $BC$. Then we have the relation:
  $$\frac r{r_a}=\tan \frac B2\tan\frac C2\ .$$


Proof: We can use the hint from the OP,
$$
\begin{aligned}
\frac r{r_a}
&=
\frac{s-a}s=
\frac{b+c-a}{b+c+a}=
\frac
{2R(\sin B+\sin C-\sin A)}
{2R(\sin B+\sin C+\sin A)}
\\
&=
\frac
{8R\;\sin \frac B2\;\sin \frac C2\;\cos \frac A2}
{8R\;\cos \frac B2\;\cos \frac C2\;\cos \frac A2}
=
\tan\frac B2\;\tan\frac C2\ .
\end{aligned}
$$
Alternatively, and for a "shorter" proof, let $A'$ be the projection of the incenter $I$ on $BC$, and $A''$ the projection of the $A$-ex-center $I_a$ on $BC$. Then $A'$ and $A''$ are symmetric to each other w.r.t. the mid point of $BC$, and considering the triangles $\Delta BA'I$ and $\Delta CA''I_a$ we have $BA'=s-b=CA''$, and thus
$$
\begin{aligned}
\tan\frac B2 &=\frac{IA'}{BA'}=\frac r{s-b}\ ,\\
\tan\frac {\pi-C}2 &=\frac{I_aA''}{CA''}=\frac {r_a}{s-b}\ ,\\[3mm]
\tan\frac B2\; \tan\frac C2
&=\frac{\tan\frac B2}{\tan\frac {\pi-C}2}
=\frac{r/(s-b)}{r_a/(s-b)}=\frac r{r_a}\ .
\end{aligned}
$$
$\square$

Now let us pass to the relation in the OP, 
$$
R_1R_3R_5=R_2R_4R_6\ ,
$$
which becomes equivalent to
$$
\color{brown}
{
\left(\tan\frac{A_{61}}2\tan\frac{A_{12}}2\right)
\left(\tan\frac{A_{23}}2\tan\frac{A_{34}}2\right)
\left(\tan\frac{A_{45}}2\tan\frac{A_{56}}2\right)
}
=
\color{blue}
{
\left(\tan\frac{A_{12}}2\tan\frac{A_{23}}2\right)
\left(\tan\frac{A_{34}}2\tan\frac{A_{45}}2\right)
\left(\tan\frac{A_{56}}2\tan\frac{A_{61}}2\right)
}
\ .
$$
$\square$
