Prove that the sum of the radii of the circles 
$ABCD$ is a cyclic quadrilateral. Prove that the sum of the radii of the circles drawn inside the triangles $\Delta ABC$ and $\Delta CDA$ is equal to the sum of the radii of the circles drawn inside the triangles $\Delta BCD$, $\Delta DAB$.

I came up with the idea that I can use the Japanese theorem to prove that the distance between the centers is equal - they make a rectangle - but I can't solve it. Please help.
 A: Hint. Use Carnot's Theorem: Given a triangle $\Delta ABC$, let $O$ denote its circumcenter, $R$ its circumradius, and $r$ its inradius. Let $O_1,O_2,O_3$ be furthermore the orthogonal projections of $O$ onto $BC, CA, AB$ respectively. We then have $$OO_1+OO_2+OO_3=R+r$$ Notice: The segment $OO_i$ is taken to be negative if $OO_i$ lies completely outside $\Delta ABC$ and positive otherwise.

Here, $\color{blue}{OO_2}$ would be negative, while $\color{red}{OO_1, OO_3}$ are positive. For convenience, let $AB=:c, BC=:a, CA=:b$. Notice that $OO_3BO_1$ is a cyclic quadrilateral since $\angle BO_3O+\angle OO_1B=90^\circ+90^\circ=180^\circ$, and hence, you can use Ptolemy's Theorem in order to infer $$\begin{align*}OB\cdot O_1O_3&=OO_3\cdot BO_1+O_3B\cdot OO_1\\\iff R\cdot \frac{b}2&=OO_3\cdot \frac{a}2+\frac{c}2\cdot OO_1\end{align*}$$ Analogously, you will obtain \begin{cases}R\cdot a=OO_3\cdot b+OO_2\cdot c\\
R\cdot b=OO_1\cdot c+OO_3\cdot a\\
R\cdot c=OO_2\cdot a+OO_1\cdot b \end{cases}
Add these up and consider the well-known equation $$r\cdot (a+b+c)=2\cdot [\Delta ABC]=OO_1\cdot a+OO_2\cdot b+OO_3\cdot c$$ (do you see now why it is important to take $OO_2$ to be negative?). The first part is just a consequence of dividing $\Delta ABC$ into three triangles with the incenter as a vertex. The second part is trivial.
$$\begin{align*}R\cdot (a+b+c)&=OO_1\cdot (b+c)+OO_2\cdot (c+a)+OO_3\cdot (a+b)\\ R\cdot (a+b+c)+r\cdot (a+b+c)&=OO_1\cdot (a+b+c)+OO_2\cdot (a+b+c)+OO_3\cdot (a+b+c)\\\iff R+r&=OO_1+OO_2+OO_3\end{align*}$$
Now, going back to your problem, it is rather easy to finish once we have this gem :)


 (I'll be referring to the image.) Notice that using Carnot's Theorem twice, once for $\Delta ABD$ and again for $\Delta BCD$, we obtain $$R+r_1=OO_1+OO_5+OO_4\qquad \text{and}\qquad R+r_2=OO_2+OO_3+OO_5$$ Notice that $OO_5$ is negative for $\Delta ABD$ and positive for $\Delta BCD$. Thus, if you add these two equations, you'll get $$r_1+r_2=OO_1+OO_2+OO_3+OO_4-2R$$ It is easy to see that this expression will be identical when referring to $r_3+r_4$.

