The circle $\omega$ is drawn through the vertices $A$ and $B$ of the triangle $A B C$. If $\omega$ intersects $A C$ at point $M$ and $B C$ at point $P$. The segment $M P$ contains the center of the circle inscribed in $A B C$. Given that $A B=c, B C=a$ and $C A=b$, find $M P$.
: Since $A M P B$ is cyclic, $\angle C M P=\angle C B A$ and $\angle C P M=\angle C A B$. Therefore the triangle $C M P$ is similar to the triangle $C B A$. Let $C M=x C B=$ $x a$, where $x$ is the coefficient of similarity. Then $C P=x C A=x b, M P=$ $x A B=x c$. If $I$ is the center of the circle inscribed in $A B C$ and $r$ - its radius, we have the following equality for the areas: $S_{C M P}=S_{C M I}+S_{C P I}$. But $S_{C M P}=\frac{1}{2} x a x b \sin C$, and $S_{C P I}=\frac{1}{2} x b r, S_{C P M}=\frac{1}{2} x a r$. Plugging these in and solving for $x$ we get $x=f r a c(b+a) r b a \sin C$. But $\frac{1}{2} a b \sin C=S_{A B C}=$ $(a+b+c) r$. Hence $x=\frac{a+b}{a+b+c}, M P=\frac{c(a+b)}{a+b+c}$.