four points lie on a circle

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Source: China south east mathematical olympiad 2008 day1 problem 3

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#1 h Jul 14, 2013, 3:22 PM • 2 Y

Y by Adventure10, Mango247

In $\triangle ABC$ , side $BC>AB$ . Point $D$ lies on side $AC$ such that $\angle ABD=\angle CBD$ . Points $Q,P$ lie on line $BD$ such that $AQ\bot BD$ and $CP\bot BD$ . $M,E$ are the midpoints of side $AC$ and $BC$ respectively. Circle $O$ is the circumcircle of $\triangle PQM$ intersecting side $AC$ at $H$ . Prove that $O,H,E,M$ lie on a circle.

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#2 h Jul 14, 2013, 4:03 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

From the right $\triangle BPC,$ we have $\angle PEC=2\angle PBC=\angle ABC$ $\Longrightarrow$ $PE \parallel AB$ $\Longrightarrow$ $PE$ is C-midline of $\triangle ABC$ $\Longrightarrow$ $M \in PE.$ Similarly $Q$ is on the line passing through midpoints $M,F$ of $AC,AB.$ $\angle QHM=\angle QPM=\angle ABD$ $\Longrightarrow$ $BAHQ$ is cyclic, i.e. $\angle BHA=\angle BQA=90^{\circ},$ i.e. $\odot(HEM)$ is 9-point circle $(N)$ of $\triangle ABC.$ $O$ is then intersection of the perpendicular bisector of $\overline{EF}$ and the perpendicular bisector of $\overline{PQ},$ which bisects $\angle FME$ externally, because $\triangle QMP$ is obviously M-isosceles $\Longrightarrow$ $O$ is midpoint of the arc $HM$ of $(N).$

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#3 h Aug 16, 2013, 3:40 PM • 1 Y

Y by Adventure10

first it's obvious $\triangle AQB = \triangle BQS$ so $QM \| BC$ ( thales ) now like this we have $EP \| AB$
let $\angle ABC = 2 \alpha$ and $\angle ACB = \gamma$
now we can write : $\angle PHM = \angle PQM = \angle PBC \Rightarrow \angle PBC = \angle PHC \Rightarrow$
$PHBC$ is cyclic.
$\Rightarrow \angle CHB = \angle CPB = 90^{0} \Rightarrow \angle CBH = 90^{0} - \gamma$
but in a right triangle if $AM$ is the median of $BC ( \angle A = 90^{0} )$ then $2AM =BC$
so : $HE = EB \Rightarrow \angle BEH = 180^{0} - ( 90^{0} - \gamma + 90^{0} - \gamma )$
$\Rightarrow \angle BEH = 2 \gamma (*)$
but $PEC = 2 \alpha (**)$
and $\angle HOM = 2 \widehat{ QH } + 2 \widehat{ QM } = 2 ( \angle MHQ + \angle HMQ ) = 2 ( \gamma + \angle QPM )$
$= 2 ( \gamma + \angle PBA ) = 2 \alpha + 2 \gamma \Rightarrow \angle HOM = 2 \alpha + 2 \gamma (***)$

$(*),(**),(***) \Rightarrow \angle MEH + \angle MOH = 180^{0} \Rightarrow$ it's done!!

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