Let $O_1, O_2$ be the centers of these two circles and $O$ be the center of the square. Let $M$ be the midpoint of $AB$. Note that $AO_1+BO_2=OC$ and $O_1, O_2$ lie on lines $AO, BO$ respectively. Lemma. $\angle O_1MO_2=90^{\circ}$. (Proof) As $O_1$ varies with constant velocity, so does $O_2$ on it's respective path. Hence, there exists a spiral similarity $O_1 \mapsto O_2$ and lines $AC \mapsto BD$. Thus, $(OO_1O_2)$ passes through a fixed point; setting $O_1=A$ and $O_2=B$ we see that $M$ is that fixed point, hence the lemma. $\blacksquare$ Draw a line $\ell$ through $M$ tangent to the circle with center $O_2$. It is clear from the lemma that it is also tangent to the circle with center $O_1$, so we are done. $\blacksquare$

Let \(C (O_1,R) , C (O_2,r) \) denote the circles inscribed in \(\angle A , \angle B\) respectively. Let their common tangent \(l\) intersect \(AB\) at \(M\), \(BC\) at \(L\) and \(DA\) produced at \( N.\) \( (O_1)\) is the excircle of \(\Delta MAN\) corres. vertex \(N\). We know that in a right angled triangle, Inradius + Exradius = Length of side on which excircle is formed. \(\Rightarrow\) Inradius of \(\Delta MAN + R =AB/2\) \(\Rightarrow\) Inradius of \(\Delta MAN = AB/2 - R = r\). Since \(\Delta MAN \sim MBL\) and their inradius are equal therefore, they are \(\cong\) and \(AM = BM\).

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 50... Sincerely Jean-Louis