Could someone tell me if there is any error in this solution? or why did something arrive that is not correct Wlog let $(ABC)$ the unit circle $O=0$ $X$ lies in $BC$: $\overline{x}=\frac{b+c-x}{bc}$ $HM_B \perp M_BX$: $\implies \frac{h-m_b}{\overline{h}-\overline{m_b}}=-\frac{m_b-x}{\overline{m_b}-\overline{x}}$ $\implies x=\frac{-ab^2+b^2c-abc+bc^2}{bc+ac-ab-a^2}$ similarly $y=\frac{-bc^2+c^2a-abc+ca^2}{ca+ba-bc-b^2}$ Then: $OXY$~$HBC \iff \frac{o-x}{o-y}=\frac{\overline{h}-\overline{b}}{\overline{h}-\overline{c}}$ (are similar in opposite orientation) $\iff \frac{x}{y}=\frac{b(a+c)}{c(a+b)}$ $\iff \frac{\frac{-ab+bc-ac+c^2}{bc+ac-ab-a^2}}{\frac{-bc+ca-ab+a^2}{ca+ba-bc-b^2}}=\frac{a+c}{a+b}$ $\iff \frac{\frac{(c-a)(c+b)}{(c-a)(a+b)}}{\frac{(a-b)(a+c)}{(a-b)(b+c)}}=\frac{a+c}{a+b}$ But this is not correct.

https://www.wolframalpha.com/input?i=%28a%2Bb%2Bc-%28a%2Bc%29%2F2%29%2F%281%2Fa%2B1%2Fb%2B1%2Fc-%281%2Fa%2B1%2Fc%29%2F2%29%3D-%28%28%28a%2Bc%29%2F2-x%29%2F%28%281%2Fa%2B1%2Fc%29%2F2-%28%28b%2Bc-x%29%2Fbc%29%29%29 I think you miscalculated the value of $x$.

WLOG, assume $AB \le AC$. Let $AD, BE, CF$ be the altitudes of $ABC$, the midpoint of $BC$ be $M_A$, the reflection of $O$ over $M_A$ be $O_A,$ the reflections of $A$ over $E, F$ be $A_1, A_2$ respectively, and $N = AB \cap O_AY$. Notice that $HDYM_C$ is cyclic with diameter $HY$ by Thales'. In addition, we have $$AH \cdot AD = |Pow_{(BDHF)}(A)| = AB \cdot AF = \frac{AB}{2} \cdot (2 \cdot AF) = AM_C \cdot AA_2$$so $HDA_2M_C$ is also cyclic. Thus, $HDA_2YM_C$ is cyclic, yielding $\angle HA_2Y = 90^{\circ}$. Now, observe $$\angle YA_2B = 90^{\circ} - \angle HA_2B = 90^{\circ} - \angle HAB = 90^{\circ} - \angle DAB = \angle DBA = \angle YBA_2$$so $YA_2 = YB$. It's well-known that $BHA_1CA_2$ is cyclic with center $O_A$. This implies $O_AA_2 = O_AB$, so $O_AY$ is the perpendicular bisector of $A_2B$. Hence, $$\angle OYX = \angle O_AYX = \angle NYB = 90^{\circ} - \angle NBY = 90^{\circ} - \angle FBC = \angle FCB = \angle HCB.$$Because an analogous process yields $\angle OXY = \angle HBC$, the desired similarity follows. $\blacksquare$ Remark: Having experience with POP within orthocenter configurations motivates the constructions of $A_1$ and $A_2$. Now, once we recall that both points also lie on $(HBC),$ it becomes very clear that $A_1$ and $A_2$ are central to this configuration. Also, at first, I didn't realize $OXY$ and $HBC$ were inversely similar, so I was trying to prove $X = BC \cap OM_B$ and $Y = BC \cap OM_C$.

Trivial. Let $C'$ be the reflection of $H$ across $M_C, H'$ be the intersection of $AH$ and $(ABC)$, $D$ is the foot of the $A$-altitude. $H'C'$ is the common chord of $C'HH'$ and $(ABC)$. Thus $OY \perp C'H' \parallel M_CD \implies \angle OYD=90-\angle M_CDB=90-B $, by symmetry $\angle OXD=90-\angle C$ which is enough to deduce the similarity.