Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.
Prove that $\angle PAB = \angle CAQ $
I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$… wait this looks like an apolonian circle.
Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.
Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$
Best Answer
Let $R$ be the intersection of $PQ$ with $AC$ and let $O$ be the circumcenter of $ABC$. Clearly, $O$ lies on the line $PQR$.
Clearly $\angle PBA = \angle PBO + \angle OBA$ and $\pi - \angle ACQ = \angle CQR + \angle QRC$. Since $\angle OBA = \frac \pi 2 - \angle ACB = \angle QRC$ and $\angle PBA + \angle ACQ = \pi$ (by assumption), it follows that $\angle PBO = \angle CQR = \angle OQB$. This proves that $\triangle OPB \sim \triangle OBQ$. This yields $OB^2=OP\cdot OQ$. As $O$ is the circumcenter of $ABC$, we have $OB=OA$ and therefore $OA^2=OP\cdot OQ$. This implies that $\triangle OPA \sim \triangle OAQ$, hence $\angle OAP = \angle OQA$. Finally, $$\angle PAB = \angle OAB - \angle OAP = \angle QRC - \angle OQA = \angle QAC.$$