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 $(N).$

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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