Here is a brute force solution. (It is not a canonical solution. I tried to
give at least a synthetic solution, but failed. The idea was to connect / to parallely project the given configuration of points (and constructed helper points) on the circle $(ABC)$ to points on the line $PA$. Since $PA$ would have been my choice of an idea to solve, it would be not canonical even if it would work...)
We will show an equivalent result. It is important to have the right order of introducing the points, to get their properties in a fluent story. As it is always the case for me, i am trying to mention and more related "coincidences" in the given geometric constellation, even with the risk of having a long(er) presentation, the "bonus results" being in my eyes always important. (The reader may want to extract the solution from the the many ingredients.) The points are constructed, and their properties are claimed in the light yellow framework, arguments are given on white background.
I was trying to use Pascal's Theorem, but failed. (When doing so for a circle with a point $A$ on it, the "line" $AA$ is the tangent in $A$ to the circle.) The problem with this idea is that it is hard to find a configuration of six points like $AA?BC?$ on the circle $(ABC)$ that is indeed useful and uses / encodes essentially the properties of $C$.
Let's go.
A rectangle $ABA'B'$ is given, let $O$ be the intersection of its diagonals $AA'$, $BB'$. We denote by $(O)$ the circumcircle of the rectangle, $(O)=(ABA'B')$. We denote by $(d)$ any direction parallel to the direction of $AB\| A'B'$. Let us denote the point at infinity on $(d)$ by $\infty=\infty_d$. (So a line contains this infinity point, iff it is parallel to $(d)$.)
Let $P$ be a point on the tangent in $A$ to $(O)$, so $PA\perp AO$.
Let $C$ be the second intersection of $PB$ with $(O)$, $C\ne B$.
Let $X$ be a further point on $(O)$.
We consider the point $E=PO\cap BX$. Let $E^*$ be the intersection $PA\cap E\infty$ (of $PA$ with the parallel through $E$ to $(d)$).
We consider the point $D=AB'\cap BX$. Let $D^*$ be the intersection $PA\cap D\infty$ (of $PA$ with the parallel through $D$ to $(d)$).
$AE$ intersects $(O)$ for a second time in a point denoted by $Y$, here $Y\ne A$.
Let $Z$ be the second point of intersection of $E^*X$ with the circle $(O)$.
Claim: We have $(d)\|YZ\|AB\|EE^*\|DD^*\|B'A'$.
Proof of the claim: Consider the hexagon $AAYZXB$ inscribed in $(O)$. Using Pascal, the points $AA\cap ZX=E^*$, $AY\cap XB=E$, $YZ\cap AB$ are colinear. So the point $YZ\cap AB$ is $\infty$, the only (projective) point on both $AB$ and $EE^*$. (Because $AB\cap EE^*=\infty_d=\infty$.) This implies $\infty\in YZ$, i.e. $YZ$ is parallel to the direction $(d)$.
Claim: The points $C,D,Z$ are colinear.
Proof of the claim:
Let us denote by $x,y,t,u$ the arc lengths for the following (marked) arcs:
- the arc from $A$ to the first intersection of $PO$ with $\overset\frown{AB}$
- the arc the first intersection of $PO$ with $\overset\frown{AB}$ to $B$, so $x+y=\overset\frown{AB}$,
- the arc $\overset\frown{BY}=\overset\frown{ZC}$,
- the arc $\overset\frown{XZ}$.
Then the Theorem of Ceva in the triangle $\Delta ABO$ w.r.t. the inner point $E$ is:
$$
\begin{aligned}
1 &=
\frac
{\sin\widehat{EAB}}
{\sin\widehat{OAE}}
\cdot
\frac
{\sin\widehat{AOE}}
{\sin\widehat{EOB}}
\cdot
\frac
{\sin\widehat{OBE}}
{\sin\widehat{EBA}}
\ ,
\\[2mm]
&\qquad\text{ or using our notations}
\\[2mm]
1 &=
\frac
{\sin(t/2)\ \cdot\ \sin(x)\ \cdot\ \sin((\pi-x-y-t-u)/2)}
{\sin(\pi-x-y-t)/2\ \cdot\ \sin(y)\ \cdot\ \sin((t+u)/2)}
\ .
\end{aligned}
$$
This is the point where we must use the important assumption on $E$, the fact that it is on $PO$.
We will show the colinearity of $C,D,Z$ by showing that $CD$ and $CZ$ is the same line. The angle $\widehat{ACB'}$ is fixed, so we take each of the lines, the sines of the angles separated by them from $\widehat{ACB'}$, and show equivalently
the relation:
$$
\tag{$*$}
\frac
{\color{gray}{AC\cdot} \sin\widehat{ACZ}}
{\color{gray}{B'C\cdot} \sin\widehat{ZCB'}}
\
\overset{(!)}{=\!\!=}
\
\frac
{\color{gray}{ AC\cdot CD\cdot} \sin\widehat{ACD}}
{\color{gray}{B'C\cdot CD\cdot} \sin\widehat{DCB'}}
\ .
$$
Note that the R.H.S. above is the quotient of the areas of the triangles
$\Delta ACD$ and $\Delta DCB'$, and we can write
$$
RHS
=
\frac
{\operatorname{Area}(ACD)}
{\operatorname{Area}(DCB')}
=
\frac{AD}{DB'}
=
\frac
{\operatorname{Area}(ABD)}
{\operatorname{Area}(DBB')}
=
\frac
{AB\cdot\sin\widehat{ABX}}
{BB'\cdot\sin\widehat{XBB'}}
\ .
$$
The relation $(*)$ is thus equivalent to
$$
\tag{$\dagger$}
\frac
{AC\cdot \sin(t/2)}
{B'C\cdot \sin((\pi-x-y-t)/2)}
\
\overset{(!)}{=\!\!=}
\
\frac
{AB \cdot\sin((t+u)/2)}
{BB'\cdot\sin((\pi-x-y-t-u)/2)}
\ .
$$
Using the Ceva relation above, this is equivalent to
$$
\tag{$\diamondsuit$}
1
\
\overset{(!)}{=\!\!=}
\
\frac{AC}{B'C}\cdot
\frac{BB'}{AB}\cdot
\frac{\sin y}{\sin x}\ .
$$
We can finally use the definition of $C$ as the point making the tangent in $A$, the line $EO$, and the line $BC$ concurent in $P$. We use for this the similarity $\Delta PAB\sim\Delta PCA$, which becomes
$\displaystyle
\frac{PC}{PA} =
\frac{PA}{PB} =
\frac{AC}{AB} $.
So we replace in $(\diamondsuit)$ the fraction $AC:AB$ by $PA:PB$.
We can finally show $(\diamond)$:
$$
\frac{AC}{AB}\cdot
\frac{B'C}{BC}\cdot
\frac{\sin y}{\sin x}
=
\frac{PA}{PB}\cdot
\frac{1}{\sin \widehat{B'BC}}\cdot
\frac{PO\cdot OB\;\sin y}{PO\cdot OA\;\sin x}
=
\frac{PA}{PB}\cdot
\frac{1}{\sin \widehat{B'BC}}\cdot
\frac
{\operatorname{Area}(POB)}
{\operatorname{Area}(POA)}
\\
=
\frac{PA}{PB}\cdot
\frac{1}{\sin \widehat{B'BC}}\cdot
\frac
{PB\cdot BO\;\sin\widehat{PBO}}
{PA\cdot AO\;\sin\widehat{PAO}}
=
\frac
{\sin \widehat{PBO}}
{\sin \widehat{B'BC}}
=1\ .
$$
$\square$
This concludes the proof.
Corollary: This concludes the problem in the OP, since
$$
\widehat{BAE} =
\widehat{BAY} =
\widehat{AYZ} =
\widehat{ACZ} =
\widehat{ACD} \ .
$$
$\square$
Unfortunately i could not make the proof more structural.
My hope was that it should be a bridge using results like Pascal's theorem and / or configurations of the shape Pappus, Desargue...
As bonus, here are some colinearities that can be established using Pascal's theorem.
The problem in the above tempting picture is that at the place where the point $D$ is placed, there are in fact three points,
- the point $D:=AB'\cap BX$,
- and the point $D_1:=CZ\cap BX$,
- and the point $D_2:=CZ\cap AB'$.
And we need the equality of two of them.
To see colinearities from the picture, apply Pascal's theorem for instance for...
- $AABXA'B'$, giving the colinearity of $D^*$, $\infty$, $D$, as mentioned above,
- $XYZCAB$, giving the colinearity of $F:=XY\cap CA$, $\infty$, $D_1$,
- $AA'XBB'Y$, giving the colinearity of $O$, $H:=A'X\cap B'Y$, $E$,
- $BCAB'YXX$, giving the colinearity of $G:=BC\cap B'Y$, $F$, $D$,
- $AYXA'CA$, giving the colinearity of $J:=AY\cap A'C$, $F$, $D^*$,
- $AYZCBA$, ...
and 
$$\angle CBF =\angle DCA
\Leftrightarrow
\frac{\sin{\angle BAE}}{\sin{\angle EAO}}
=\frac{\sin{\angle ACD}}{\sin{\angle DCH}}
=\frac{AD*HC}{DH*AC}$$
$$\frac{\sin{\angle BAE}}{\sin{\angle EAO}}
=\frac{AD*CH*\sin{\angle EOB}}{DH*BA*\sin{\angle AOE}}$$
It only need to prove that:
$$\frac{AB*HC}{BH*AC}
=\frac{\sin{\angle EOB}}{\sin{\angle AOE}}
=\frac{\sin{\angle OPB}*\sin{\angle PAB}}{\sin{\angle OPA}*\sin{\angle PBA}}$$
But I have no idea about it. Any ideas? 
