A hard question about plane geometry

COMMENT.-There are many triangles that satisfy the initial conditions and not all of them verify the possibility that $H$ is in the incircle. The cases in which this is true, it seems to me that are very few and even perhaps just one (I mean all the answers could be similar).

Anyway, all these kind of triangles $\triangle{ABC}$ can be defined by the points $B=(2a,0),(-a,0), C=(0,2b),(0,-b)$ which determines the point $A=(-2a,-2b)$. From this, the circumcenter can be deduced from the system $$\begin{cases}b(y+b)=-2ax\\2by=-a(x+a)\end{cases}\implies (x_O,y_O)=\left(\frac{a^2-2b^2}{3a},\frac{b^2-2a^2}{3b}\right)$$ Similarly one has for the orthocenter $$(x_H,y_H)=\left(\frac{4b^2-2a^2}{3a},\frac{4a^2-2b^2}{3b}\right)$$ On the other hand the coordinates of the incenter are given by $$x_I=\frac{2a(\sqrt{a^2+4b^2}-\sqrt{a^2+b^2})}{\sqrt{a^2+b^2}+\sqrt{4a^2+b^2}+\sqrt{a^2+4b^2}}\\y_I=\frac{2b(\sqrt{4a^2+b^2}-\sqrt{a^2+b^2})}{\sqrt{a^2+b^2}+\sqrt{4a^2+b^2}+\sqrt{a^2+4b^2}}$$ With the side $\overline{AC}$ whose line have equation $2bx-ay+2ab=0$, we have for the inradius $r$ $$r=\frac{|2bx_I-ay_I+2ab|}{\sqrt{a^2+b^2}+\sqrt{4a^2+b^2}+\sqrt{a^2+4b^2}}$$ Parameters $a,b$ giving suitable triangles can be taken from the equalities $$\overline{IH}=\overline{IO}=r$$ Calculations become less arduous if we make $a=1$ (so for each value of the resultant $b$, all similar triangle is available).

In picture N is the center of nine point circle and is the mid point of OH. Now all we have to show is that $ON\bot BP$ Wich Will prove that O is also on in circle.The data that $FC\bot BE$ deduces that :

$arc BC+arc QS=180^o$

Now the rotation of construction $QC\bot BS$ about vertex B such that :

1- S coincides on P

2- Q coincides on U

must result in the coincidence of C on M which in turn results in $UM\bot BP$ which deduces that the diameter UM, which HO locates on bisects the chord BP. In this way BP is perpendicular bisector of the chord HO that says O is also on the incircle.