Orthocenter of a triangle collinear with two points in the circumcircle. 

This difficult elementary geometry problem was proposed by @Nyafh54 and receiving no answer was deleted twice despite several upvotes. We republish it here mainly for the information of the O.P. who showed be interested in the subject, and those who gave their opinion in favor of the problem.
The statement is true even for shapes of the triangle such that, for example, the segment $DOE$ is not contained inside the triangle. We have this, in particular, for the triangle of vertices $A=(3,4),B=(0,0), C=(8,0)$ in the attached figure.
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HINT.-$(1)$ Vertices $A;B;C$ given, we know how to determine the circumcenter $O$, the orthocencer $H$ and the circumcircle.
$(2)$ $P$ and $S$ points are determined by the bisector of $BC$ side and $D$ and $E$ points by the bisector of segment $AS$ so we have got three points $D,E,S$.
$(3)$ The statement can lead people to the construction of the circumcircle of the triangle $\triangle{DES}$ then by intersection with the first circumcircle to determine the $X$ point and this is the difficulty of the problem for beginners.
$(4)$ This difficulty can be avoided by intersecting the line $PH$ with the first circumcircle so we get a point $X$. We don’t know yet that $X$ belongs to the circumcircle of new triangle $\triangle {DES}$ but we can show this proving that the quadrilateral $DESX$ is cyclic. For this, already having the coordinates $(x_i,y_i)$ of the points $D,E,S,X$ we can verify the Ptolemy's theorem: $$\overline{DE}\cdot\overline{XS}+\overline{DX}\cdot\overline{ES}=\overline{DS}\cdot\overline{XE}$$ Another way is using the equation of the circumcircle of $\triangle {DES}$
$$\det\begin{vmatrix} x^2+y^2&x&y&1\\x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\end{vmatrix}=0$$ and verify this equality  putting the fourth point $X$ instead of the generic $(x.y)$.
 A: 
Let M be the intersection of $AS$ and $PX
$. Clearly $P,O,S$ lies on the same line, which is a diameter of circle $O$, so we have $\angle PXS=90^\circ$. Now since $AS$ bisects $\angle BAC$ and $DE$ perpendicularly bisects $AS$, quadrilateral $AESD$ is a rhombus and $DS=ES$, so we know $MS$ lies on a diameter line (symmetry line) of the blue circle. Furthermore since $\angle MXS=90^\circ$ we know point $M$ lies on the blue circle.
Even further, since $\angle MED = \angle MSD = \angle MAD$, and also since $AM\perp ED$, we know $EM\perp AD$ and $M$ is the orthocenter of $\triangle AED$.

Forget about the whole $PX$ line first. Denote the other intersection of $AD$ and the blue circle as $T$. Denote the other intersection of $AE$ and the blue circle as $U$.
Construct $H$ as the orthocenter of $\triangle ABC$. Since the details are pretty long, here's a sketch of the essential steps:
(1) The intersection $CH$ and $ST$, denoted $R$, lies on circle $O$. The intersection of $BH$ and $SU$, denoted $V$, also lies on circle $O$.
(2) $H$ lies one the line $TU$.
(3) $XH$ bisects $\angle TXU$ from angle bisector theorem on ${TX\over UX}={TH\over UH}$.
(4) Therefore $X,H,M$ lies on the same line, which is the original $PX$ line.
Here's the details:
(1) First notice that $D,E,T,U$ are cocyclic on the blue circle, $\triangle ATU$ is similar to $\triangle AED$ which is isosceles. So $T,U$ are symmetric over the diameter $M,S$. Next, since $EM$ is parallel to $CH$ (both are perpendicular to $AB$), we know $$\angle AEM= \angle ACR$$ Also since $M,E,U,S$ are cocyclic on the blue circle, $\angle AEM=\angle ASU$. Applying the symmetry of $T,U$ over $MS$, we have $\angle ASU=\angle AST=\angle ASR$. So overall we have $$\angle AEM=\angle ASR$$ This means $\angle ACR = \angle ASR$ and therefore $R$ lies on the circumcircle of $\triangle ACS$, which is circle $O$. $V$ lies on circle $O$ goes by the exact same logic.
(2)Since $S$ is the midpoint of $BC$, $VU$ bisects $\angle CVH$. Also since $\angle ASR=\angle ASV$ we have arc $AR$ and $AV$ are the same arc, so $CU$ bisects $\angle VCH$. Therefore $U$ is the incenter of $\triangle VCH$ and $HU$ bisects $\angle VHC$. By similar argument $T$ is the incenter of $\triangle RBH$ and $HT$ bisects $\angle RHB$. So $HT$ and $HU$ lies on the same line which is $TU$.
(3)

Notice that arcs $RX$ and $TX$ corresponds to the same angle $\angle RSX$ in two circles, while arcs $VX$ and $UX$ also corresponds to the same angle $\angle VSX$ in the same two circles, so $\triangle RXV$ and $\triangle TXU$ are similar. From this similarity we obtain $${XT\over XU}={XR\over XV}$$
In addition, since $\angle RXV=\angle TXU$ which means $\angle RXT=\angle VXU$ and the above ${XT\over XU}={XR\over XV}$, we have triangles $\triangle RXT$ and $\triangle VXU$ are similar too. From this similarity we obtain $${XR\over XV}={RT\over UV}$$
Combining the two emphasized results we get $${XT\over XU}={RT\over UV}$$

Now go back to this diagram, since $\triangle RBH$ and $\triangle VCH$ are similar for obvious reason, and $T,U$ are the corresponding incenters of the two triangles, we get $\triangle RTH$ and $\triangle VUH$ are also similar and $${RT\over UV}={TH\over UH}$$
Combining all emphasized results above, we get ${XT\over XU}={TH\over UH}$ which completes our angle bisector theorem condition for step (3). Indeed $XH$ bisects $\angle TXU$.
(4) $XH$ bisects $\angle TXU$ means $XH$ bisects $\angle DXE$ as well, because arcs $DT$ and $EU$ are the same angle. This means the extension of $XH$ lies on $M$, which is the midpoint of arc $DE$. This implies $X,H,M$ are colinear, which completes our whole proof.
