Since AO, BO are isogonals of AH, BH in the triangle $\triangle ABC$, O is its circumcenter. AN is a circumcircle diameter, hence $NM \perp AM$ and $NM \parallel BC$. M is a refelection of the orthocenter H in the side BC, hence E is the midpoint of HM. BC is the midline of the right angle triangle $\triangle HNM$ parallel to its side NM, hence P is the midpoint of HN. OP connects the the midpoints O, P of the hypotenuses AN, HN of the right angle triangles $\triangle ANM, \triangle HNM$, hence $OP \perp NM$ and $OP \perp BC$. Thus P is also the midpoint of BC and $AH \equiv AM \parallel OP \equiv OS$. BC is the perpendicular bisector of the base HM of the trapezoid HMSO passing through its diagonal intersection R, which means that this trapezoid is isosceles and P is also the midpoint of its other base OS. As a result, the diagonals HN, OS of the quadrilateral HNSO cut each other in half at P and it is a parallelogram, i.e., $AN \equiv ON \parallel HS$.

i found one of these days that the problem was still true if $ H$ and $ O$ were substituted by an arbitrary internal point $ P$ and its isogonal conjugate (if that's the name) of $ P$. i have a solution with vectors, but it comes to be a bit long and tired, does anybody have a better solution?

campos wrote: i found one of these days that the problem was still true if H and O were substituted by an arbitrary internal point P and its isogonal conjugate (if that's the name) of P. Indeed, the result is still true for two isogonal conjugates WRT ABC. Proposition: Let $ P$ and $ Q$ be two isogonal conjugates WRT $ \triangle ABC.$ Rays $ AP,AQ$ cut its circumcircle $(O)$ at $P',Q'.$ Define the points $ M \equiv BC \cap QP',$ $N \equiv PQ' \cap BC$ and $ L \equiv PM \cap QN.$ Then $ APLQ$ is a parallelogram. $ \triangle A'B'C'$ is pedal triangle of $ P$ WRT $ \triangle ABC.$ Thus, $ QA \perp B'C',$ $QB \perp C'A'$ $\Longrightarrow$ $ \angle AQB = 180^{\circ} - \angle A'C'B' = 180^{\circ} - (\angle PBA' + \angle PAB') = 180^{\circ} - \angle PBP'$ $ \frac {\overline{P'M}}{\overline{MQ}} = \frac {\overline{BP'}}{\overline{BQ}} \cdot \frac {\sin \widehat{P'BC}}{\sin \widehat{CBQ}} \ \ , \ \ \frac {\overline{P'P}}{\overline{PA}} = \frac {\overline{BP'}}{\overline{BA}} \cdot \frac {\sin \widehat{PBP'}}{\sin \widehat{ABP}}$ By sine law for $ \triangle ABQ$ we get: $ \frac {\overline{BA}}{\overline{BQ}} = \frac {\sin \widehat{AQB}}{\sin \widehat{BAQ}}$ Combining with the two latter expressions, keeping in mind that $ \angle AQB = 180^{\circ} - \angle PBP',$ $\angle BAQ = \angle P'BC$ and $ \angle CBQ = \angle ABP,$ we get $ \frac {_{\overline{P'M}}}{^{\overline{MQ}}} = \frac {_{\overline{P'P}}}{^{\overline{PA}}}$ $ \Longrightarrow$ $ PM \parallel AQ.$ Mutatis mutandis, $ QN \parallel AP.$

Let $P'$ be the feet of perpendicular dropped from $O$ to $BC$. It's known that $2OP=AH$ Hence, $\frac{OP}{AH}=\frac{OP}{AN}$ $\implies H,P',N$ collinear.Hence, $P=P'$ So, $OP \parallel AH$ . It suffices to prove $OP=PS$ Since $HE=EM$ the result follows quite easily.