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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 $MD=PQ$.

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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 $DH=MP$?

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

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  • $\begingroup$ Thank you for the response! I don't get what happens after $PQ=MQ-MP$. $\endgroup$ – Stellar Sep 28 '19 at 14:44
  • $\begingroup$ 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. $\endgroup$ – Hw Chu Sep 28 '19 at 17:06
  • $\begingroup$ Got it. Thank you! May I ask why $\dfrac{1}{2}(BM-AM)=DM$? $\endgroup$ – Stellar Sep 28 '19 at 17:27
  • $\begingroup$ Since $D$ is the midpoint of $\overline{AB}$, $\overline{BM} = \overline{BD} + \overline{DM}$ and so on. $\endgroup$ – Hw Chu Sep 28 '19 at 17:30
  • $\begingroup$ Can you continue it for me? I really messed up. $\endgroup$ – Stellar Sep 28 '19 at 17:31
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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 $AD = BD.$

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