Show that all lines XY pass through point H We are given a circle $\Gamma$ centered at $O$ and a line $r$ (in red) outside of it.
Point $A$ is the feet of $O$ in $r$.
Let's take an arbitrary point $P$ in the circumference of $\Gamma$ and let's build circle $\omega$ whose diameter is $PA$.
Let $X$ and $Y$ be the second meetings of $\omega$ with $\Gamma$ and $r$, respectively.
Show that as $P$ varies along $\Gamma$ line $XY$ will have a fixed point $H$.

Point $A'$ is the inverse of $A$ wrt to circle $\Gamma$.
Point $Z$ is the second intersection of $XY$ with $\Gamma$
My first atempt was to show that $ZOPA$ is cyclic but I couldn't do it on my own.
 A: 
Since $OA' \times OA = OP^2$ and $\angle A'OP = \angle POA$, $\triangle OA'P \sim \triangle OPA$. Then,
$\angle OPA' = \angle OAP = 90° - \angle PAY = 90° - \angle PXY$
$= 90° - \angle XPZ - \angle XZP = 90° - \frac{\angle POZ}{2} = \angle OPZ = \angle OZP$
It implies $PA'Z$ is a straight line and $OPAZ$ is a cyclic quadrilateral.

Let $M$ be the other intersection point of $OA$ and the circle $AXPY$. Joint $PM$ and $PY$.
$\angle PMA = \angle PYA = \angle YAM = 90°$
$\Rightarrow AMPY$ is a rectangle
$\Rightarrow PM$ is equal and parallel to $YA$
Let $P'$ be the other intersection point of $PM$ and the circle $PXZ$. Also, delete the line which is parallel to $YA$ and passes through $A'$.
$\triangle P'AM \cong \triangle PAM$
$\Rightarrow \angle P'AM = \angle PAO = \angle PZO = \angle ZPO = \angle ZAO$
$\Rightarrow ZP'A$ is a straight line.

When $MA \lt A'A$ and $OM \lt OP$,

Joint $A'P'$, $A'X$, $P'X$ and $MX$.
$\angle A'P'X = \angle PP'X + \angle PP'A' = \angle PP'X + \angle P'PZ = \angle PP'X + \angle P'XC = \angle P'CZ = \angle AYZ = \angle AMX$
$\Rightarrow A'MXP'$ is a cyclic quadrilateral.
Let $N$ be the other intersection point of $ZY$ and the circle $A'XP'M$. Joint $A'N$.
$\Rightarrow \angle A'NH = \angle XMA = \angle AYH$
$\Rightarrow A'N$ is parallel to $YA$.
$\Rightarrow \angle NHA' = \angle YHA$ and $NA' = P'M = PM = YA$
$\Rightarrow \triangle NA'H \cong \triangle YAH$
$\Rightarrow A'H = AH$
$\Rightarrow$ Since $A'$ and $A$ are fixed points, $H$ is a fixed point.

When $MA \ge A'A$ and $OM \lt OP$,

Let $C$ be the intersection point of $PP'$ and $YX$. Joint $A'P'$, $A'X$, $P'X$ and $MX$.
$\angle A'P'X = \angle PP'X - \angle PP'A' = \angle PP'X - \angle P'PZ = \angle PP'X - \angle P'XC = \angle P'CZ = \angle AYZ = \angle AMX$
$\Rightarrow A'XP'M$ is a cyclic quadrilateral.
Let $N$ be the other intersection point of $ZY$ and the circle $A'XP'M$. Joint $A'N$.
$\Rightarrow \angle A'NH = \angle A'MX = \angle AYH$
$\Rightarrow A'N$ is parallel to $YA$.
$\Rightarrow \angle NHA' = \angle YHA$ and $NA' = P'M = PM = YA$
$\Rightarrow \triangle NA'H \cong \triangle YAH$
$\Rightarrow A'H = AH$
$\Rightarrow$ Since $A'$ and $A$ are fixed points, $H$ is a fixed point.
