Bariz - Problem

Source: 2021 Cono Sur Shortlist G5 https://artofproblemsolving.com/community/c1088686_cono_sur_shortlist__geometry

Tags: geometry, equal angles

parmenides51

01.11.2022 01:52

Let $\vartriangle ABC$ be a triangle with circumcenter $O$, orthocenter $H$, and circumcircle $\omega$. $AA'$, $BB'$ and $CC'$ are altitudes of $\vartriangle ABC$ with $A'$ in $BC$, $B'$ in $AC$ and $C'$ in $AB$. $P$ is a point on the segment $AA'$. The perpenicular line to $B'C'$ from $P$ intersects $BC$ at $K$. $AA'$ intersects $\omega$ at $M \ne A$. The lines $MK$ and $AO$ intersect at $Q$. Prove that $\angle CBQ = \angle PBA$.

Since $M$ lies on the circumcircle and on $AH$, its isogonal conjugate must lie on the line at infinity and on $AO$. This gives implies that the reflection of $BM$ in the $B$-anglebisector is parallel to $AO$. Let the point at infinity along $AO$ be $\infty_{AO}$. Now by DDIT on $\mathcal{Q}\{AP,PK,KQ,QA\}$ we get an involution swapping $(BA,BK)$, $(BP, BQ)$, $(BM,B\infty_{AO})$ but since $BA$, $BK$ and $BM$, $B\infty_{AO}$ are reflections of each other in the $B$-angle bisector, it must be that $BP$ and $BQ$ are reflections of each other in the $B$-angle bisector so we are done.