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Source: 2018 China North Mathematical Olympiad Grade 10 Test 2 P2

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#1 h Jul 26, 2018, 1:46 AM • 1 Y

Y by Adventure10

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 $\angle ABM=\angle ACM$

This post has been edited 2 times. Last edited by Devastator, Aug 14, 2018, 12:46 AM
Reason: Edit

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#2 h Jul 26, 2018, 2:08 AM • 3 Y

Y by Mr_pi, Adventure10, Mango247

hint

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#3 h Jul 26, 2018, 2:59 PM • 2 Y

Y by Adventure10, Mango247

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, $\angle ABM = \angle ABH + \angle HBM = \angle ACH + \angle HCM = \angle ACM$ .

This post has been edited 1 time. Last edited by Man0308Fong, Jul 26, 2018, 3:02 PM

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Plops

#4 h Mar 15, 2020, 5:57 AM • 2 Y

Y by teomihai, Mango247

Thanks to tworigami for the diagram:

https://i.imgur.com/kuiteQm.png

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, $\angle ABM=\angle AHM=\angle MCA$ .

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