# 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 $$DH=MP$$?

• I noticed that $PQ=BI-AG$. I'm not sure if that is useful. – Daniel Mathias Sep 28 '19 at 13:18

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 $$\overline{DM}$$.

• Thank you for the response! I don't get what happens after $PQ=MQ-MP$. – Stellar Sep 28 '19 at 14:44
• This is a fact for circles inscribed in triangles. In your diagram, $\overline{MQ} = \overline{MH} =$ half perimeter minus the opposite side ($\overline{BC}$). Similarly $\overline{BH} = \overline{BI} =$ half perimeter minus the opposite side ($\overline{CM}$). To prove this, try writing down equations and solve them. – Hw Chu Sep 28 '19 at 17:06
• Got it. Thank you! May I ask why $\dfrac{1}{2}(BM-AM)=DM$? – Stellar Sep 28 '19 at 17:27
• Since $D$ is the midpoint of $\overline{AB}$, $\overline{BM} = \overline{BD} + \overline{DM}$ and so on. – Hw Chu Sep 28 '19 at 17:30
• Can you continue it for me? I really messed up. – Stellar Sep 28 '19 at 17:31

Since the points $$EMDH$$ occur in that sequence along a line, we know that $$DE + DH = EM + HM.$$

If you can show that $$DE - DH = HM - EM$$ then from these two equations you can conclude that $$DH = EM = MP.$$

From $$PQ = MQ - MP$$ you can easily find that $$HM - EM = PQ.$$

Showing that $$DE - DH = PQ$$ takes more steps (at least the way I did it). You can start with $$PQ = CP - CQ$$ but then you have to chase differences of segment lengths all the way around the outside of the triangle, using the facts that $$AC = BC$$ and $$AD = BD.$$