Geometry problem involving orthocenter and collinearity Let $X$, $Y$ on line $BC$ from triangle $ABC$ such that $\angle XAY = 90^\circ$. Suppose that $H$ is the orthocenter of $ABC$. If $AX$ intersects $BH$ at $X'$ and $AY$ intersects $CH$ at $Y'$, prove that intersection of circumcircle of $BXX'$ and circumcircle of $CYY'$ lay on $X'Y'$

I showed that $\angle BEX' + \angle CEY'= 90$ so showing $\angle BEC = 90^\circ$ would do the trick so we can do something like showing $BFCE$ cyclic or $BFGE$ cyclic.
Am I going in the right direction or is there something else we need to do or some theorem we need to use?
Please give any hint you can.
 A: Lemma: Circumcircles $AXY$, $BXX'$, $CYY'$ are concurrent at point $D$.
Corollary: Where $E$ is the second intersection of circumcircles $BXX'$ and $CYY'$, show that $\angle BEC = 90^ \circ$. Hence the desired result follows.

 $\angle BEC = 360^\circ - \angle BED - \angle CED = 360^\circ - (180^\circ - \angle BXD    ) - (180^\circ - \angle CYD ) = \angle CYD + \angle YXD = 180^\circ - \angle XDY = 90^\circ$.

Proof of Lemma: We will use the following theorem (stated without proof).
Theorem: Miquel's quadrilateral theorem:


*

*Given 3 straight lines, a circle may be drawn through their intersections (IE Circumcircle)

*Given 4 straight lines, the 4 circles so determined through sets of 3 points, meet in a point.

This is proved by applying Miquel's theorem, which is just angle chasing.

Consider the lines $AXX', X'BH, AYY', XBCY$.
Then, Miquel shows that circumcircles $AXY$, $BXX'$, $AIY'$ and $AJX'$ are concurrent.
Let this point be $D$.
Consider the lines $AYY', Y'CH, AXX', XBCY$.
Again, Miquel shows that circumcircles $AXY$, $CYY'$, $AIY'$ and $AJX'$ are concurrent, which is point $D$.
Hence, the 3 circumcircles are concurrent at $D$.

Note:

*

*Showing that the 3 circles are concurrent seemed to be the crux. In fact, there are many circles (EG circumcircle of $ BGFC$) which pass through this point.

*However, it was hard to work with these 3 circles directly (because $BXX'$ and $CYY'$ are "too far apart"), which is why we defined other circles (that helped us bridge the gap).

*I wish there was a nicer way to show that the 3 circles are concurrent. At it's heart, it's just an angle chasing proof. I'm still thinking of how we could do this (that isn't just applied Miquel).

A: 
The proof given below is an angle-chasing exercise, which mostly uses properties of cyclic quadrilaterals. Although it does not necessarily invoke Miquel Theorems, one can resort to use a certain property of Miquel Point and Miquel Triangle as the very last argument to finish off the proof.
Before attempting to affirm OP’s statement, we need to prove the following lemma. It provides us a lead in form of a point, which is pivotal to the said proof.
$\underline{\mathbf{Lemma}}$
The scalene triangle $ABC$ and the right angle triangle $AXY$ share the vertex $A$. Besides, the sides $BC$ and $XY$ of these two triangles lie on the same straight line. If $H$ is the orthocenter of $\triangle ABC$, prove that the circumcircles of $\triangle AXY$ and $\triangle HBC$ meat on the altitude of $ABC$ that passes through its vertex $A$.
$\underline{\mathbf{Proof\space of\space the\space Lemma}}$
Assume that $Q$ is the point of intersection between the circumcircle of $\triangle AXY$ and the altitude of $ABC$ that passes through its vertex $A$. Draw the lines $BQ$ and $CQ$ as shown in $\mathrm{Fig.\space 1}$.
Since $H$ is the orthocenter of $\triangle ABC$, $AGHF$ is a cyclic quadrilateral.
$$\therefore\quad \measuredangle CHB = \measuredangle GHF = 180^o - \measuredangle CAB. \tag{1}$$
Since $AXY$ is a right angle triangle, $XY$ is a diameter of its circumcircle $\Gamma_1$ . Since $AQ$ is a chord of this circle, $AQ$ is perpendicular to $XY$. Furthermore, $P$, the point of intersection between $XY$ and $AQ$, bisects the latter.
The two right angle triangles $BPA$ and $QPB$ are SAS-congruent. Therefore, we have
$$AB=BQ. \tag{2}$$
The two right angle triangles $APC$ and $CPQ$ are also SAS-congruent. Thus, we have
$$AC=CQ. \tag{3}$$
The two triangles $ABC$ and $BQC$, which share a side (i.e. $BC$), are SSS-congruent because of (2) and (3). Therefore, we shall write,
$$ \measuredangle BQC = \measuredangle CAB. \tag{4} $$
From (1) and (4), it is obvious that
$$ \measuredangle BQC + \measuredangle CHB = 180^o,$$
which means that $HBQC$ is a cyclic quadrilateral.
The circumcircle of $HBQC$ happens to be the circumcircle of the triangle $HBC$. Therefore, the circumcircles of $\triangle AXY$ and $\triangle HBC$ meat at $Q$, which lies on the altitude of $ABC$ passing through its vertex $A$.

$\underline{\mathbf{Proof\space of\space OP’s\space Statement}}$
Please note that we have changed the labeling of the points $X'$ and $Y'$ of the OP's diagram to $M$ and $N$ respectively.
$\mathrm{Fig.\space 2}\space$ shows an extended version of the configuration provided by OP. We have added the circumcircle $\mathit{\Gamma}_4$ of $\triangle HBC$ and the altitude $AP$, which runs through the vertex $A$ of $\triangle ABC$. We have proved above that the circumcircle $\mathit{\Gamma}_1$ of $\triangle AXY$ intersects with $\mathit{\Gamma}_4$ at a point that lies on $AP$. Let us denote it and the other point of intersection between the two circumcircles as $Q$ and $D$ respectively. We construct the three lines $DA$, $DH$ and $DQ$. For brevity, we let
$$\measuredangle DQA = \phi.$$
The names of the circumcircle of $\triangle BXM$ and its point of intersection with the line $MN$ are $\mathit{\Gamma}_2$ and $E$ respectively. Finally, we add the six lines $DB$, $DC$, $DX$, $DY$, $DM$, and $DN$. Our aim is to prove that $E$ lies on the circumcircle of $\triangle NYC$ as well.
Since the two exterior angles $\measuredangle DXM$ and $\measuredangle DBM$ were formed by producing one of the sides of the cyclic quadrilaterals $DQAX$ and $DQHB$ respectively, they both are equal to the respective interior opposite angles $\measuredangle DQA$ and $\measuredangle DQH$. Therefore, we have
$$\measuredangle DXM = \measuredangle DBM = \phi, \tag{5}$$
which means that the line $MD$ subtends equal angles at $X$ and $B$ on the same side of it. Consequently, we can state that a circle can be drawn through the four points $X$, $B$, $M$, and $D$ or, in other words, the circle $\mathit{\Gamma}_2$ passes through $D$.
Since $\measuredangle DEM$ and $\measuredangle DBM$ are in the same segment of the circle $\mathit{\Gamma}_2$,
$$\measuredangle DEM = \measuredangle DBM = \phi. \tag{6}$$
Since $\measuredangle NED$ is a supplement of $\measuredangle DEM$, we shall write
$$\measuredangle NED = 180^o - \phi. \tag{7}$$
Since $\measuredangle DCH$ and $\measuredangle DQH$ are in the same segment of the circle $\mathit{\Gamma}_4$, they are equal, i.e.
$$\measuredangle DCH = \measuredangle DQH = \phi. \tag{8}$$
In similar vein, $\measuredangle DYA$ and $\measuredangle DQA$ are equal because they are in the same segment of the circle $\mathit{\Gamma}_1$.
$$\measuredangle DYA = \measuredangle DQA = \phi. \tag{9}$$
We name this juncture of our proof as $\bf\it{\space J\space}$ for us to refer to it in the note given at the end of this text.
Since $\measuredangle NCD$ and $\measuredangle NYD$ are supplements of $\measuredangle DCH$ and $\measuredangle DYA$ respectively, using (8) and (9) we can write
$$\measuredangle NCD = \measuredangle NYD = 180^o - \phi. \tag{10}$$
Due to similar reasons mentioned above, we can maintain that point $D$ lies on the circumcircle $\mathit{\Gamma}_3$ of  $\triangle CNY$, which is not shown in the diagram.
We can infer from (7) and (10) that
$$\measuredangle NED = \measuredangle NCD.$$
This means that $E$ lies on the circumcircle of $\triangle NYC$, because the two equal angles $\measuredangle NED$ and $\measuredangle NCD$ are subtended by the same segment $ND$.
$\underline{\mathbf{\color{red}{Note}}}$
According to Miquel’s (Triangle) Theorem, the three lines drawn from Miquel Point of a triangle, say $\mathbf{\triangle}$,  to the vertices of the Miquel Triangle make equal angles with the respective sides of $\bf{\triangle}$. At the juncture $\space \bf{\it{J}}\space$ of our proof, we have the necessary information to use this property of Miquel Point to conclude the proof.
A: Here is yet "an other" solution, i am trying to show as many "coincidences" (that can be extracted from a picture) in the  given constellation of points. This is a slightly changed version of the problem in the OP, but in the final stage taking the picture from the OP and the picture from the present reformulation, then pasting them point after point together, we see that the two pictures coincide. The point $E$ from the OP will be constructed a priori in a different way, so to make clear the difference, i will use the notation $E'$ for the new point. (Its final properties show it is $E$ from the OP.) Some points are introduced in the same way as in the OP. But $D,X',Y'$ will be introduced in a different way (without changing notations this time). I hope that the transposition / glueing between the the two statements is clear. So let us restate (and add a lot of new points and a lot of new properties)...

Proposition:
Four points $X,Y;B,C$ are given on a line. Let $A$ be a point on the circle with diameter $XY$. To fix some notations used in the proof, let $x,y$ be the corresponding angles in $X,Y$ in $\Delta AXY$, $x+y=90^\circ$. Let also $\xi$, $\eta$ be the angles $\widehat{XAB}$ and $\widehat{CAY}$ as in the picture below.
We construct the heights $AA_1$, $BB_1$, $CC_1$ in $\Delta ABC$, their intersection is its orthocenter $H$. We construct the intersections of lines $X'=AX\cap BHB_1$ and
$Y'=AY\cap CHC_1$.
Let $E'$ be the projection of $A$ on $X'Y'$.
Then:

*

*$(1)$ $AB_1E'X'$ cyclic.

*$(1')$ $AC_1E'Y'$ cyclic.

*$(2)$ $BC_1B_1CE'$ cyclic.

*$\color{blue}{(3)}$ $\color{blue}{X,X',B,E'}$ are on a circle, let us denote it by $\Gamma_X$.

*$\color{blue}{(3')}$ $\color{blue}{Y,Y',C,E'}$ are on a circle, let us denote it by $\Gamma_Y$.

(This already concludes the OP. Furthermore...) $E'$ is thus one point of intersection of the circles $\Gamma_X$
and $\Gamma_Y$, let $D$ be the second point of intersection.
Let $B'$ be the second point of intersection of $AB$ with $\Gamma_X$.
Let $C'$ be the second point of intersection of $AC$ with $\Gamma_Y$.
We also construct the intersections $Y_1=B'X\cap Y'CHC_1$, $X_1=C'Y\cap X'BHB_1$.
Then we have the bonus properties:

*

*$(4)$ $X'DC'$ is a line.

*$(4')$ $Y'DB'$ is a line.

*$(5)$ $AX_1YE'B$ cyclic.

*$(5')$ $AY_1XE'C$ cyclic.

*$(6)$ $HBCD$ cyclic.

*$(7)$ $X_1HE'Y'$ cyclic.

*$(7')$ $Y_1HE'X'$ cyclic.



Above, there is an obvious "symmetry" of the pairs of properties, obtained by exchanging $X,B\leftrightarrow Y,C$, so e.g. $(1)$ and $(1')$ claim "the same".
Proof:
$(1)$ $\hat B_1=\hat E'=90^\circ$ (in the two triangles with hypothenuse $AX'$).
$(2)$ The points $B,C,B_1,C_1$ are on the circle with diameter $BC$. Let us show that $E'$ is on the same circle. We compute:
$$
\begin{aligned}
\widehat{B_1E'A} &\overset{(1)}=
\widehat{B_1X'A} = 90^\circ-\widehat{B_1AX'} =
90^\circ-\xi-\hat A=\eta\ ,\\
\widehat{C_1E'A} &\overset{(1')}=\xi\ ,\qquad\text{by symmetry, so}\\
\widehat{B_1E'C_1} &=\xi+\eta =90^\circ-\hat A=\widehat{B_1BC_1}\ , 
\end{aligned}
$$
so $E'BC_1B_1$ cyclic.
$(3)$
$$
\begin{aligned}
\widehat{BE'C_1} &\overset{(2)}=
\widehat{BCC_1} = 90^\circ-\hat B\ ,
\\
\widehat{C_1E'A} &=\xi\qquad\text{ as above ,}
\\
\widehat{BE'A} &= \widehat{BE'C_1} + \widehat{C_1E'A} = 90^\circ-\hat B+\xi\ ,\\
\widehat{BE'X'} &= 90^\circ-\widehat{BE'A}
=\hat B-\xi
\\
&=(180^\circ -\hat A-\eta-y)-\xi\qquad\text{ from }\Delta BAY
\\
&=x\qquad\text{ from }\Delta XAY
\\
&=180^\circ-\widehat{BXX'}\ , 
\end{aligned}
$$
so $BXX'E'$ cyclic. This proves the OP.
$\square$

Let us show also the further claimed bonus points.
$(5a)$ Above, we have obtained $\widehat{BE'X'}=x$, so
$$
\widehat{BE'A}=90^\circ-x=y=\widehat{BYA}\ ,
$$
so $ABE'Y$ cyclic. (We show soon that a further point on the corresponding circle, the point $Y_1$, to get $(5)$.) Similarly, we get  $(5'a)$, $ACE'X$ cyclic.
$(4)$ We compute the angle $\widehat{X'DC'}$ as the sum of the two angles obtained from it after the dissection with the half-line $DE'$. So let us compute:
$$
\begin{aligned}
\widehat{E'DC'} &\overset{\Gamma_Y}=
180^\circ - \widehat{E'CC'} = \widehat{E'CA} \ ,
\\
\widehat{E'DX'} &\overset{\Gamma_X}=
180^\circ - \widehat{E'XX'} = \widehat{E'XA} \ ,
\\
\widehat{E'DC'} + 
\widehat{E'DX'}
&=
\widehat{E'CA} +
\widehat{E'XA} \overset{(5'a)}= 180^\circ\ ,
\end{aligned}
$$
proving $(4)$.
$(5)$ We have already seen in $(5a)$ that $ABE'Y$ is cyclic, let us also add the point $X_1$. For this we compute the angle...
$$
\begin{aligned}
\widehat{BX_1Y} &=
\widehat{X'X_1C'} =
180^\circ-\widehat{X_1X'C'} -\widehat{X_1C'X'}
\\
&= 
\underbrace{180^\circ-\widehat{B_1X'C'} -\widehat{B_1C'X'}}_{=\hat B_1=90^\circ}-\widehat{X_1C'B_1}
=90^\circ - \widehat{YC'C}
\\
&=90^\circ - \widehat{YC'C}
=90^\circ - \widehat{YY'C}
=90^\circ - \xi
\\
&=\widehat{BAY}
\ ,
\end{aligned}
$$
showing $(5)$. Similarly, $(5')$ is obtained.
$(6)$
$$
\begin{aligned}
\widehat{BDC} &= 
\widehat{BDE'}+\widehat{E'DC}
=
\widehat{BX'E'}+\widehat{E'Y'C}
\\
&=
180^\circ-\widehat{X'HY'}
=
180^\circ-\widehat{BHC}
\ ,
\end{aligned}
$$
showing $(6)$.
$(7)$
$$
\widehat{HX_1E'} =
\widehat{BX_1E'} \overset{(5)}=
\widehat{BYE'}  =
\widehat{CYE'}  =
\widehat{CY'E'}  =
\widehat{HY'E'}  
\ .
$$
This concludes $(7)$ and all bonus points.
$\color{green}\blacksquare$
