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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?
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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

Math stackexchange 3822556

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 $PA=PA'$.

$\square$

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