Let $ABC$ be an isosceles triangle ($AB=AC$). An arbitrary point $M$ is chosen on the extension of the $BC$ beyond point $B$. Prove that the sum of the radius of the circle inscribed in the triangle $AMB$ and the radius of the circle tangent to the side $AC$ and the extensions of the sides $AM, CM$ of the triangle $AMC$ does not depend on the choice of point $M$.
Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(b,0)$; choose $M(-\lambda,0)$.
The bisector of $\angle AMB\ :\ ax-y(\lambda+u)+a\lambda=0$ (with $u=\sqrt{a^{2}+\lambda^{2}}$) cuts
the bisector of $\angle ABM\ :\ y=-\frac{b+w}{a}(x+b)$ (with $w=\sqrt{a^{2}+b^{2}}$) in the point $O_{1}$
and $y_{O_{1}}=r_{1}=\frac{a(\lambda-b)(b+w)}{a^{2}+(b+w)(b+v)}$.
The bisector of $\angle AMB\ :\ ax-y(\lambda+u)+a\lambda=0$ cuts the external bisector
of $\angle ACB\ :\ ax+y(b-w)-ab=0$ in the point $O_{2}$ and $y_{O_{2}}=r_{2}=\frac{a(b+\lambda)}{b+\lambda+v-w}$.
Then $r_{1}+r_{2}=a$.