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Let $ABC$ be a triangle. Let the external bisector of angle $A$ meet the circumcircle of triangle $ABC$ again at $M \neq A$. A circle with centre $M$ and radius $MB$ meets the internal bisector of angle $A$ at points $P$ and $Q$. Determine the length of $PQ$ in terms of the lengths of $AB$ and $AC$.

Could anyone please provide a solution? I cannot seem to make any significant progress in the question.

Edit: Here is the original project that I created in Geogebra. Hope it makes the diagram clearer.


  • $\begingroup$ Nice formatting of question; nice diagram. Re "...I cannot seem to make any significant progress...", please edit your query to show all of the work that you have done so far. $\endgroup$ Aug 22, 2020 at 8:24
  • $\begingroup$ It is basic angle chasing and most of it is just me making sense of the diagram. And a big mess of trigonometric expressions. I do not want to include this as I want an elegant answer to the question. $\endgroup$ Aug 22, 2020 at 8:30
  • $\begingroup$ I regard your response as reasonable. However, the first thing that mathSE reviewers will focus on is that you haven't shown your work. A mathSE reference on this issue is math.meta.stackexchange.com/questions/9959/…. Also, please use mathJax to format your math. A mathSE reference on this is math.stackexchange.com/help/notation. $\endgroup$ Aug 22, 2020 at 8:38
  • $\begingroup$ I am very sorry, but typing that out would be a nightmare, especially to a novice like me. And I do use Mathjax (same as Latex, right?) $\endgroup$ Aug 22, 2020 at 8:49
  • $\begingroup$ Mathjax has some minor differences with Latex. Use <br> to force line break. Also, Mathjax can only be invoked by enclosing the characters in $...$ or $$...$$. Thus, for example to underline text via mathJax rather than html tags, you have to do something like $\underline{\text{abc}}$. $\endgroup$ Aug 22, 2020 at 8:55

1 Answer 1


Try to prove this..

•Find the length of $MA=2R\cos(\frac{A+2C}{2})$ first. ( Where $R$ is the circumradius of the triangle.)

•Then find $MB=2R\cos(\frac{A}{2})$ by using Sine Law (Chase the angles) in $\triangle MAB$

•Finally apply Pythagoras theorem in $\triangle MAQ$

$MQ^2-MA^2=MB^2-MA^2=AQ^2$ and $PQ=2AQ$

  • $\begingroup$ But what is Angie AMB? $\endgroup$ Aug 22, 2020 at 9:09
  • $\begingroup$ It is equal to angle ACB. $\endgroup$
    – LM2357
    Aug 22, 2020 at 9:11
  • 1
    $\begingroup$ Oh sorry I am an idiot $\endgroup$ Aug 22, 2020 at 9:12
  • $\begingroup$ Well I did get an expression but it is not pretty. Something horrific in terms of sines of angles B and C $\endgroup$ Aug 22, 2020 at 9:19
  • $\begingroup$ Can you post what did you get? When I was doing it, I replaced $\frac{b}{\sin B}=\frac{c}{\sin C}=2R$...See if this helps. $\endgroup$
    – LM2357
    Aug 22, 2020 at 9:21

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