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.
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$\begingroup$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 $PQ=2AQ$
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