How to prove that $\angle{APC}+\angle{AYC}=180$? 
Let $ABC$ be an acute triangle and $E,F,G$ be the altitude on $AB,BC,AC$ respectively and $H$ be an orthocentre. $D$ is the intersecting point of $EF$ and $BH$ and $P$ is the mid-point of $BH$. Extended $BH$ at $Y$ such that $\angle{ADC}=\angle{AYC}$. Prove that $APCY$ is a cyclic.

I let $X, Z$ be the intersecting point of $AP,CD$ and $CP,AD$ respectively. To prove that $APCY$ is a cyclic I wanted to prove that $APC+AYC=180$. Since, $\angle{ADC}=\angle{AYC}=\angle{XDZ}$ I wanted to prove that $\angle{XPZ}+\angle{XDZ}=180$. I noticed that $X,Z$ in on the circle of cyclic $AEFC$ so than $\angle{PXD}+\angle{DZP}=180$. But how could I prove that $X,Z$ is on the circle?
Sorry for the inconvenience without photo. I don't know how to add or draw the geometry in here. I hope you could help me with this. Thank you ❤️
 A: I will try to give two solutions. The first solution is a direct synthetic solution, for this i need a simple lemma, then use inversion. The second solution is a prosaic solution, added in order to show what can be done analytically in short straightforward computations in such cases, and although such a solution breaks the beauty, such a way to go may be useful in mathematical competitions. 
Finally, observe that one can give also an "inverse solution" in the sense that we construct points $X'$, $Z'$ instead of $X$, $Z$ as in the comment of Mick, showing afterwards that $X=X'$, $Z=Z'$. (From the alternative construction, the properties used are better suited to be combined with the given situation.) 
Let's go.

Direct solution: We need the following...

Lemma: Let $\Delta ABC$ be a triangle, let $H$ be its orthocenter, let $O$ be the center of the circumscribed circle, and we denote this circle by $(O)$ or $(ABC)$.

On the line $BH$ we consider the following points: $G$, its intersection with $AC$; $D^*$, its intersection with the circumcircle $(O)$; and $P^*$, the reflection of $B$ with respect to $G$, i.e. $BG=GP^*$. On the line $CH$ we consider $E$, its intersection with $AB$.
Then the line $AP^*$, the line $CD^*$, and the parallel to the line $BHGD^*P^*
$  through $E$ intersect in a point, $X^*$. In other words, the reflection w.r.t. line $AGC$ maps the points $B,E,H$ respectively in $P^*,X^*, D^*$.
In particular, the trapezoid $AEX^*P^*$ is isosceles, thus cyclic. 
The four points $B,F,D^*,X^*$ are on a circle.

Proof of the lemma: We have
$$
\widehat{HCA}=\frac \pi2-\hat A=
\widehat{HBA}=
\widehat{D^*BA}=
\widehat{D^*CA}\ .
$$
The points $B,H$ are thus reflected in $D^*,P^*$, and since $A,C$ are invariated by the reflection, the intersection $E=AB\cap CF$ is reflected in $X^*=AP^*\cap CD^*$. Then $EX^*$ is perpendicular on the axis of reflection, as $AP^*$ is, too, so $EX^*\|AP^*$. To see $BFD^*X^*$ cyclic, we test if there is a same power of $C$ with respect $B,F$, and with respect to $X^*,D^*$, and indeed
$$
CF\cdot CB=CH\cdot CE=CD^*\cdot CX^*\ .
$$
$\square$
Now we place the OP in the foreground, and restate and show, recalling and enriching the data of the problem:

Let $\Delta ABC$ be a triangle with orthocenter $H$.
  The heights of this triangle are (denoted in a slightly unusual manner) $AF$, $BG$, $CE$, with $E,F,G$ on the sides of the triangle. Let $D$ be the point of intersection $EF\cap BH$. Let $P$ be the mid point of the segment $BH$. Let $X$ be the intersection $AP\cap CD$. Let $Z$ be the intersection $CP\cap AD$.
We also introduce the inversion denoted by a star, $W\to W^*$, centered in $B$ and with power
  $$
BE\cdot BA=BH\cdot BG=BF\cdot BC\ .
$$
  Then we have the following:
  
  
*
  
*(1) $A=E^*$, $E=A^*$; $G=H^*$, $H=G^*$; $F=C^*$, $C=F^*$.
  
*(2) $D^*$ (the image of $D$ by this inversion) is the intersection of the line $BHG$ with the circumcircle $(ABC)$.
  
*(3) $P^*$ is the symmetric point of $B$ w.r.t. $G$. (Because of $2BP=BH$, and $H^*=G$.)

  
  
*
  
*(4) $X^*$ is the intersection point of $BX$, $AP^*$, $CD^*$, and the circle $(BXP^*)$. Moreover, $CX^*\perp AX^*$.
  
*(5) The quadrilaterals $(BXDF)=(\infty X^*D^*C)^*$, $(BPXE=\infty P^*X^*A)^*$ are cyclic.
  
*(6) The angles in $X,Z$ in the quadrilateral $XPZD$ are right angles.
  
*(7) Let $Y$ be the symmetric point of $D$ w.r.t. to the point reflection in $G$, then $APCY$ is cyclic. 

Proof: The above points were collected for an easy structure of the proof. Most of them are clear when stated, we will only touch the hard points. Note that the inversion is well defined, since the power of $B$ w.r.t. the circles $GHEA$, $GHFC$ give the equality $BE\cdot BA=BH\cdot BG=BF\cdot BC$. Now (1), (2), (3) are immediate. 
(4) follows from the previously isolated lemma. Indeed, $X$ is characterized by being on the lines $\infty AP$ and $\infty CD$, so after applying the inversion $X^*$ is characterized by being on $(\infty AP)^*=(BEP^*)$ and $(\infty CD)^*=(BFD^*)$.
So this is exactly the point denoted by coincidence $X^*$ in the lemma, and it also lies on $BX$, $AP^*$, $CD^*$. 
(5) is clear, the inversion of a line is a circle through the center of inversion.
(6) uses (4),
$$
\begin{aligned}
\widehat{PXD}
&=
\widehat{BXD}
-
\widehat{BXP}
\\
&=
\widehat{BD^*X^*}
-
\widehat{BP^*X^*}
\\
&
=
\widehat{D^*X^*P^*}
=
\widehat{CX^*P^*}
=90^\circ\ .
\end{aligned}
$$
(7)
$$
\widehat{AYC}+
\widehat{ADC}
=
\widehat{APC}+
\widehat{XDZ}
=
180^\circ\ .
$$
$\square$

Note: The picture comes with further smog, showing some bonus relations, one can show "the other" related properties to obtain "the other" (corresponding) solution. The essence was isolated in the lemma.

Solution by computation: We show that $GA\cdot GC=GP\cdot GY$. (A posteriori, we can affirm as a matter of terminology  that $G$ has this same power in the cyclic quadrilateral $(APCY)$. Showing the relation is a proof of the cyclicity.) 
Let us denote by $A,B,C$ the (measures of the) angles in $\Delta ABC$, by $a,b,c$ the (lenghths of the) corresponding opposite sides. Let $R$ be the radius of the circumcircle. Then we have:
$$
\begin{aligned}
GA &= c\cos A=2R\; \sin C\cos A\ ,\\
GC &= a\cos C=2R\; \sin A\cos C\ ,\\
GA\cdot GC &= 4R^2 \; \sin A\cos A\; \sin C\cos C\ ,\\[2mm]
GP &= \frac 12(GB+GH)\\
&=\frac 12(c\sin A+AG\underbrace{\tan\widehat{HAG}}_{\cot C})\\
&=\frac 12(2R\; \sin A\sin C+2R\; \sin C\cos A\cdot\frac{\cos C}{\sin C})\\
&=R\cos(A-C)\ ,\\
GD & = GB - DB\\
&=c\sin A -\frac{BE\cdot BA}{BD^*}\\
&=2R\;\sin A\sin C -\frac{a\cos B\cdot c}{2\cdot R\sin\frac {\widehat{BOD^*}}2}\\
&=2R\;\sin A\sin C -2R\;\frac{\sin A\cos B\cdot \sin C}{\sin \widehat{BAD^*}}\\
%&=2R\;\sin A\sin C\Big(\ 1 -\frac{\cos B}{\sin (A+\widehat{CAD^*})}\ \Big)\\
&=2R\;\sin A\sin C\Big(\ 1 -\frac{\cos B}{\cos(A-C)}\ \Big)\ .
\end{aligned}
$$
At the last passage we have used 
$\widehat{BAD^*}
=A+\widehat{CAD^*}
=A+\widehat{CBD^*}
=A+(90^\circ-C)$, so the sine of this angle is the cosine of $(A-C)$. This implies
$$
\begin{aligned}
GP\cdot GD 
&= 2R^2\; \sin A\sin C\; \Big(\ \cos(A-C) -\cos B\ \Big)\\
&= 2R^2\; \sin A\sin C \cdot (-2)\sin\frac{A-C+B}2\sin \frac{A-C-B}2\\
&= -4R^2\; \sin A\sin C \;\sin\frac{180^\circ-2C}2\sin \frac{2A-180^\circ}2\\
&= 4R^2\; \sin A\sin C \;\cos A\cos C\\
&= GA\cdot GC\ .
\end{aligned}
$$
$\square$
A: I think you have solved the problem; you correctly showed that $\angle PXD+ \angle PZD=180^o$. PG is the hight of triangle PAC from vertex P, also PX and AZ cross D which is on this hight. All thses indicate that CX and AZ are also hights of triangle PAC. That is:
$\angle PXD= \angle PZD=90^o$
In this case quadrilateral PXDZ is cyclic. Therefore:
$\angle XPZ+ \angle ZDX=180^o$
But:
$\angle ZDX= \angle ADC=\angle AYC$
⇒$\angle APC+ \angle AYC=180^o$
And this means quadrilateral AYCP is cyclic. 

