Let $H$ be orthocenter of an acute $\Delta ABC$, $M$ is a midpoint of $AC$. Line $MH$ meets lines $AB, BC$ at points $A_1, C_1$ respectively, $A_2$ and $C_2$ are projections of $A_1, C_1$ onto line $BH$ respectively. Prove that lines $CA_2, AC_2$ meet at circumscribed circle of $\Delta ABC$. Proposed by Anton Trygub
Problem
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Tags: geometry
IMO2021
14.05.2019 22:11
Denote by $R$ the projection $B$ onto $MH$. It is well-known that $R \in (ABC)$. Then $BRC_1C_2$ is cyclic, so $\angle RC_2C_1 = \angle RBC_1 = \angle RBC = \angle RAC$, so $R, C_2, A$ are collinear since $C_1 C_2 \parallel AC$, analogiously $R, A_2, C$ are collinear and we are done.
Aiden-1089
26.03.2024 12:29
Let $X$ be the $B$-Queue point, $H'_B$ be the second intersection between $BH$ and $(ABC)$, $B'$ be the antipode of $B$ in $(ABC)$. It is well-known that $X,H,M,B'$ are collinear, and $AH'_BB'C$ is an isosceles trapezium with $AC // H'_BB'$. We claim that $X$ lies on $CA_2$ and $AC_2$. Since $\measuredangle BXA_1=90^{\circ}=\measuredangle BA_2A_1, BXA_1A_2$ is concyclic. $\measuredangle B'XC = \measuredangle ABH'_B = \measuredangle A_1BA_2=\measuredangle A_1XA_2=\measuredangle B'XA_2$. So $X, C, A_2$ are collinear. Similarly, $X$ lies on $AC_2$, so we are done. $\square$