I am trying to look deeper into the problem that I posted earlier - Inscribed circles . Let me give you it again:
Triangle $\triangle ABC$ is an isosceles triangle. Point $D$ is the midpoint of $AB$, and $M$ is lying on $AD$. Circle $k_1(O_1;r_1)$ is inscribed in $\triangle AMC$ and touches $CM$ in $P$. Circle $k_2(O_2;r_2)$ is inscribed in $\triangle BMC$ and touches $CM$ in $Q$. Show that $MD=PQ$.
Is it possible to find $AM: BM$ if we know $PQ=\dfrac{3}{7}AB$? If $PQ=3x$, then $AB=7x$ and $AM+BH+DH=4x$.
