Prove that 4 points are concyclic in a shape which includes right-angle triangles, bisectors, and circumcircles of triangles I have come across the following question:
Let $\triangle ABC$ be right-angled at $A$ and let $AE \perp BC$. Let $Z\neq A$ be a point on the line $AB$ with $AB=BZ$, 
$(c)$ the circumcircle of $\triangle AEZ$, 
$D$ the second point of intersection of $(c)$ with $ZC$,
$F$ the antidiametric point of $D$ with respect to $(c)$, 
$P=FE\cap CZ$. 
If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T, E, B, Z$ are concyclic

I made the following observations:

*

* $\angle EZD=\angle EFD$ (as $DEFZ$ is cyclic) 

* $\angle AZE=\angle AFE$ ($AFZE$ cyclic) 

*$\angle FED=90^\circ$ ($FD$ diameter)

*$B$ is the point where the perpendicular bisector of $AZ$ from $O$ intersects $AZ$

*$\angle BAC=\angle AEC=90^\circ$, and $\angle ACE$ is common,$\implies \triangle BAC\sim \triangle AEC$

*Radii $OZ=OD \implies \triangle OZD$ is isosceles

*We also have that $\angle OAB=\angle AZT$

This is all I managed to think of for this question. Is there anyway to solve it based on some or all of my observations?
 A: The idea is to show that the points $Z,B,E,P,T$ lie on the circle with diameter $ZT$.
I will start with a nice picture serving for illustration of the solution

and a not so nice comment. Composer of geometry problems often take a series of ten points introduced in a simple way, they get a concrete geometric constellation, then they find for the simplest points a (most) complicated way to introduce them. Each simple property becomes hard, except for the case when we also reverse the order. Terminology: "Composition by contorsion". Example: The point $P$ is simply the projection of $A$ on $CZ$. I will use $P'$ from the beginning for this projection, in a final it turns out that $P=P'$. (The "contorsion" is best combined with the choice of all letters from alphabet, so that the reader finds it hard to remember them and their properties.) This is a good way to produce "hard problems" e.g. for challenges, but is not a good way to structurally educate the young geometric eye and give it a direction of study.

I will break the solution into pieces. (Still keeping the notations for the parallel.)
Lemma:
In the right triangle $\Delta ACZ$, $\hat A=90^\circ$, let $CB$ be the median. Let $E$ be the projection of $A$ on $BC$. Set $R=AE\cap CZ$. Let $AP'$, $RL$, $CE$ be the heights in $\Delta ARC$, which intersect in its orthocenter $H$.
Then: $Z,E,L$ are colinear,  $\widehat{ZER}=\widehat{ACZ}=\widehat{REP'}$, and $ER$ bisects $\widehat{ZEC}$.
Proof: From $RL\|ZA$ (right angles formed with $AC$), the reciprocal of the theorem of Ceva applied in $\Delta AZC$ starting from
$$
\underbrace{
\frac{LA}{LC}\cdot
\frac{RC}{RZ}}_{=1}\cdot
\underbrace{\frac{BZ}{BA}}_{=-1} = -1
$$
gives the concurrence of the cevians $CB$, $AR$, $ZL$. We consider the angles now and observe the relations:
$$
\widehat{ZER}
=
\widehat{EAZ}+
\widehat{EZA}
=
\widehat{ACB}+
\widehat{BCZ}
=
\widehat{ACZ}
\ .
$$
Indeed, $\widehat{EAZ}=\widehat{EAB}=90^\circ-\hat B= \widehat{ACB}$. The other equality of angles follows from the similarity $\Delta BZE\sim\Delta BCZ$. There is a common angle in $B$, and
$$
\frac{BZ}{BC}=
\frac{BE}{BZ}
$$
because of $BZ^2=AB^2=BC\cdot BE$. (The similitude "inside $\Delta BCZ$" is obtained by the similitude "inside $\Delta ACB$".)
$\square$

We come back to our problem.
Let $O$ be the center of $(c)$. The reflection in $O$ will be denoted by a star, so it is the map $X\to X^*$. For example, $F=D^*$.
Then $ACZA^*$ is a parallelogram. (Since $\widehat{CAZ}=90^\circ=\widehat{CAZ}$. Its diagonals intersect in $B$.) We know the two angles in $A^*$ in this parallelogram formed by the diagonal $ABCE$ with two sides, same as in $C$, so $AEZA^*$ cyclic, so $A^*$ is also on the circle $(c)$, its center $O$ is the mid point of $AA^*$ (because of the right angle in $E$).
Also $F=D^*$ is the point making $ADA^*F$ a parallelogram.
Let $Z^*$ be ($Z$ reflected in $O$). Then $ZAZ^*A^*$ is also a rectangle, and $AZ^*\|BO\|ZA^*$, and obtain $AZ^*=ZA^*=CA$.
Let us show that $F,E,P'$ are colinear. We show $\widehat{ACZ}=\hat C=\widehat{REP'}$. (The colinearity follows since $AER$ is a line.) We compute:
$$
\widehat{AEF} =
\widehat{AZF} =
90^\circ-\widehat{AZC} =
\widehat{ACZ} = \hat C
\widehat{REP'} 
\ .
$$
Let $\gamma$ be the circle $\gamma = \odot(EBZ)$. Its center is denoted by $O'$.
Because of
$$
\frac 12\overset\frown{BZ}=
\widehat{BEZ} =
\widehat{A^*EZ} =
\widehat{A^*AZ} =
\widehat{Z^*ZA} =
\widehat{OZA}
$$
the line $OZ$ is tangent in $Z$ to $\gamma$. So the two circles $c,\gamma$ intersect orthogonally in $Z$ (and $E$). The point $P=P'$ is also on $\gamma$ because of
$$
\widehat{EPZ} =
\widehat{EPR} =
\widehat{RAC} =
\widehat{EAC} =
\widehat{CBA} =
\widehat{EBA} 
\ ,
$$
so $PEBZ$ cyclic.
Finally, since $\widehat{ZPA}$ is a right angle, the point $T$ making $ZT$ a diameter of $\gamma$ is on $PA=PA'$.
$\square$
