Let $H$ be the orthocenter of triangle $ABC$. Let $D$ and $E$ be points on $AB$ and $AC$ such that $DE$ is parallel to $CH$. If the circumcircle of triangle $BDH$ passes through $M$, the midpoint of $DE$, then prove that $\angle ABM=\angle ACM$
Problem
Source: 2018 China North Mathematical Olympiad Grade 10 Test 2 P2
Tags: geometry, China, circumcircle
26.07.2018 17:59
Let $P = BH \cap AC, Q = CH \cap AB$. $H$ is the orthocenter of $\Delta ABC$ $\Rightarrow$ $CH \perp AB , BH \perp AC$. $DE \parallel CH$ $\Rightarrow$ $DE \perp AB$ $\Rightarrow$ $\angle BDM = 90^{\circ}$. $B,D,H,M$ are concyclic $\Rightarrow$ $\angle BHM = \angle BDM = 90^{\circ}$ $\Rightarrow$ $BH \perp HM$ $\Rightarrow$ $HM \parallel AC$ ($HM \parallel CE$). $CH \parallel DE$ $\Rightarrow$ $CH \parallel ME$ $\Rightarrow$ $CHME$ is a parallelogram. $\Rightarrow$ $DE = 2DM = 2CH$. $CH \parallel DE$ $\Rightarrow$ $CH$ is one of the middle line of $\Delta PDE$ $\Rightarrow$ $C$ is the middle point of $PE$. $M$ is the middle point of $DE$ $\Rightarrow$ $CM$ is another middle line of $\Delta PDE$ $\Rightarrow$ $CM \parallel DH$. $CH \parallel DM$ $\Rightarrow$ $CHDM$ is a parallelogram $\Rightarrow$ $\angle HDM = \angle HCM$. $B,D,H,M$ are concyclic $\Rightarrow$ $\angle HBM = \angle HDM = \angle HCM$. $CH \perp AB, BH \perp AC$ $\Rightarrow$ $CQ \perp AB, BP \perp AC$ $\Rightarrow$ $\angle BQC = \angle BPC = 90^{\circ}$ $\Rightarrow$ $B,Q,P,C$ are concyclic $\Rightarrow$ $\angle ABH = \angle ACH$. Therefore, $\angle ABM = \angle ABH + \angle HBM = \angle ACH + \angle HCM = \angle ACM$.
15.03.2020 08:57
Thanks to tworigami for the diagram: Let $BH \cap BC=P$, $CH \cap BC=Q$ Then $CQ || ED$ and $\angle DBH=\angle HME=\angle HCA=\angle HCE$ since $\triangle ACQ ~\triangle ABP$, so $HMCE$ is a parallelogram. Similarly, since $CH ||DM$ and $DM=ME=CH$, $CMDH$ is a parallelogram. Therefore, $\angle ABM=\angle AHM=\angle MCA$.