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 $MD=PQ$.

enter image description here

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 $DH=MP$?

  • $\begingroup$ I noticed that $PQ=BI-AG$. I'm not sure if that is useful. $\endgroup$ – 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 $\overline{DM}$.

  • $\begingroup$ 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

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 $AD = BD.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.