# 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

• $$\angle EZD=\angle EFD$$ (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?

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 $$PA=PA'$$.

$$\square$$