A segment divides a square into two polygons of with unequal incircles ... $1.$ Point A and B lie on the sides of a square, segment AB divides the square into two
polygons each of which has an inscribed circle. One of the circles has radius 6 cm
while the other one is larger. What is the difference, in cm, between the side length
of the square and twice the length of segment AB ? 
What must a polygon meet in order for it to have a inscribed circle? I am having trouble in just finding the right figure to work with in this problem, I think I'll have no problem finding the answer after the figure, but complete solutions are welcome
 A: Refer to the diagrams below. We focus on the bottom right quarter of the square.

The square has a side length $$2CM=CM+CN$$.
Now,
$$=\begin{align*}\\&\implies CM+CN-2AB\\&\implies CM+CN-(AP+BP)-AB\\&\implies CM+CN-(AN+BM)-AB\\&\implies(CM-BM)+(CN-AN)-AB\\&\implies 
BC+AC-AB\\&\implies BC+AC-(AQ+BQ)\\&\implies
BC+AC-(AE+BD)\\&\implies CD+CE=12\end{align*}$$
Note that we applied equal tangent segments repeatedly.
A: Let the incircles have centers $R$ and $S$ and respective radii $r$ and $s$ (with $r < s$). Note that $s$ is half the side of the original square. Let $T$ be the the original square's vertex that completes right triangle $\triangle ABT$, and define $c:= |AB|$.

Zooming in ...

The four pairs of congruent right triangles allow us to write (naming squares by their diagonals)
$$\begin{align}
|\square ST| - |\square RT| &= 2\,|\triangle ABS| + 2\,|\triangle ABR| \tag{1} \\[6pt]
s^2 - r^2 &= 2\cdot \frac12 sc + 2\cdot\frac12 rs \tag{2} \\[6pt]
(s+r)(s-r) &= c(s+r) \tag{3} \\[6pt]
s-r &= c \tag{4}
\end{align}$$
The relation in the original question is merely $(4)$, "doubled". $\square$
