# Find the length of PQ. 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.

https://www.geogebra.org/classic/ezted9sg

• 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. Aug 22, 2020 at 8:24
• 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. Aug 22, 2020 at 8:30
• 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. Aug 22, 2020 at 8:38
• 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?) Aug 22, 2020 at 8:49
• 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 $\underline{\text{abc}}$. Aug 22, 2020 at 8:55

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 $$PQ=2AQ$$

• But what is Angie AMB? Aug 22, 2020 at 9:09
• It is equal to angle ACB. Aug 22, 2020 at 9:11
• Oh sorry I am an idiot Aug 22, 2020 at 9:12
• Well I did get an expression but it is not pretty. Something horrific in terms of sines of angles B and C Aug 22, 2020 at 9:19
• 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. Aug 22, 2020 at 9:21