Q:

Inscribed circles

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 $MD=PQ$.

We have equal tangent segments: $AE=AG, CG=CP,ME=MP,MH=MQ,BH=BI,CQ=CI$. 
$MH=MQ$, thus $MD+DH=MP+PQ$
How can I prove $DH=MP$?

A:

You just have not used the fact that $\triangle ABC$ is isosceles.
$$
\begin{aligned}
\overline{PQ} &= \overline{MQ} - \overline{MP}\\
&= \frac12\left(\overline{BM}+\overline{MC}-\overline{BC}\right) - \frac12\left(\overline{AM}+\overline{MC}-\overline{AC}\right)\\
&= \frac12\left(\overline{BM}-\overline{AM}\right).
\end{aligned}
$$
And this is $\overline{DM}$.

