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Let BC be the shortest side of $\triangle$ABC. Let P be a point in AB such that $\angle$PCB=$\angle$BAC and Q be a point in AC such that $\angle$QBC=$\angle$BAC. Prove that the line that passes through the circumcenters of $\triangle$ABC and $\triangle$APQ is perpendicular to BC.

What I´ve tried so far is to show that in O is the intersection of BQ and CP, then $\triangle$OBC is isosceles with BQ=CP. then the line that passes through the circumcenter of $\triangle$ABC and O is the bisector of BC. Then if the circumcenters of $\triangle$APQ, $\triangle$ABC and O are collinear, the exercise follows.

Also,if the circumcircles of $\triangle$ABC and $\triangle$APQ intersect at two points, it also suffices tho show that the intersection of its diagonals and the circumcenters are collinear, because that means that the line that passes through the intersection of the circumcircles and BC are parallel.

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Let $R$ be the another point (different of $P$) where the $CP$ meets the circumcircle of $APQ$, let $F$ be the point where $BQ$ and $CP$ meets. Since $A$, $P$, $R$ and $Q$ lies in the same circumference it follows $$\measuredangle CRQ =180^{\circ}-\measuredangle PRQ=\measuredangle PAQ=\measuredangle BAC$$ Also, $\measuredangle BAC=\measuredangle PCB=\measuredangle RCB$ for construction, so $\measuredangle CRQ=\measuredangle RCB$ and $QR$ is parallel to $BC$. Then $\angle QRF=\angle FCB=\angle BAC=\angle QBC =\angle FQR$, i.e. $\triangle FBC$ and $\triangle FQR$ are isosceles. It follows that the bisector of $BC$ is perpendicular to $BC$ and passes by $F$, the same for the bisector of $QR$, so the bisector of $BC$ is the bisector of $QR$ too, and it is the line by the circumcircles of $\triangle ABC$ and $\triangle BPQ$

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  • $\begingroup$ Thank you so much $\endgroup$ – user670565 May 20 at 0:10

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