Junior Balkan Mathematical Olympiad 2020- P2

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cyclic quadrilateral

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Source: JBMO 2020

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250 posts

#1 h Sep 11, 2020, 1:14 PM • 4 Y

Y by dangerousliri, adityaguharoy, ItsBesi, Mango247

Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ to $BC$ . Let $Z \ne A$ be a point on the line $AB$ with $AB = BZ$ . Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$ . Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$ . Let $P$ be the point of intersection of the lines $FE$ and $CZ$ . If the tangent to $(c)$ at $Z$ meets $PA$ at $T$ , prove that the points $T$ , $E$ , $B$ , $Z$ are concyclic.

Proposed by Theoklitos Parayiou, Cyprus

This post has been edited 2 times. Last edited by Lukaluce, Sep 11, 2020, 6:44 PM

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i3435

#3 h Sep 11, 2020, 1:55 PM • 2 Y

Y by albgeo, yayitsme

Sol

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58 posts

Steff9

#4 h Sep 11, 2020, 1:56 PM • 1 Y

Y by Lukaluce

Solution

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Reason: hide text

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925 posts

#5 h Sep 11, 2020, 5:29 PM • 1 Y

Y by ItsBesi

This problem was proposed by Theoklitos Parayiou, Cyprus.

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#9 h Sep 11, 2020, 6:41 PM • 1 Y

Y by Mango247

I think the initial configuration was inspired by the humpty point so that's why there's the redundant information about $AB=BZ$ . A more general configuration (while not similar much to the first one) is this:
Let $Z$ be the reflection of $A$ over $B$ in triangle $ABC$ . Also let $E$ be the point on $CB$ such that $\angle AEB = \angle BAC$ (in the problem it is given they are both $90^{\circ}$ ). Let $(c)$ be the circumcircle of $\triangle AEZ$ and let it cut sides $CA$ and $CB$ at $P$ and $Q$ .Let the circumcircle of $AEC$ cut $BC$ at $X$ . Let the circumcircle of $BEC$ cut $CA$ at $Y$ .
(Interesting facts: $E$ is the humpty point of triangle $AEZ$ , $XY,PQ,AZ$ concur.)
(Small note: $X,Y,E$ aren't collinear, it just looks like that in the picture, sorry )

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#10 h Sep 12, 2020, 7:55 AM

Y by

My solution
Note that $<ACB=<ZAE=<ZFE=<EZT=<EDP=x$ and that $<FZP=AEB=90$ . Therefore $<ABE=<FPZ=90-x$ . Thus $PEBZ$ cyclic (1). In addition, $<CAE=90-x=<EPD$ . Thus $CAEP$ is cyclic. And therefore, $<ACE=<APE=<EZT=x$ , which means that $PEZT$ is cyclic (2). Combining (1) and (2), proves the claim.

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#11 h Sep 12, 2020, 1:15 PM • 1 Y

Y by ItsBesi

My solution which is similar to other solutions. $\angle ABE=90-\angle BAE=90-\angle ZAE=90-\angle ZFE=90-\angle ZFP$ and $90-\angle ZFP=\angle FPZ=\angle EPZ$ , since $\angle FZP=\angle FZD=90$ $\implies$ $\angle ABE=\angle EPZ$ $\implies$ $ZPEB$ is cyclic. We also have that $\angle CAE=90-\angle BAE=\angle ABE=\angle EPZ$ $\implies$ $\angle CAE=\angle EPZ$ $\implies$ $ACPE$ is cyclic $\implies$ $\angle ACE=\angle APE$ , but $\angle ACE=\angle ACB=90-\angle ABC=90-\angle ABE=\angle BAE=\angle EZT$ $\implies$ $\angle APE=\angle EZT$ $\implies$ $EPTZ$ is cyclic. Since $ZPEB$ and $EPTZ$ are both cyclic $\implies$ $T, E, B, Z$ are concyclic. $\blacksquare$

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#12 h Sep 12, 2020, 1:55 PM

Y by

Gist of what I wrote during exam

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#13 h Sep 13, 2020, 4:23 AM • 3 Y

Y by Mango247, Mango247, Mango247

Here is my solution

I always like to simplify the problem and start from the new form. Start with some angle chasing

$\angle{ZPF}=90^{\circ}-\angle{ZFP}=90^{\circ}-\angle{ZFE}=90^{\circ}-\angle{ZAE}=90^{\circ}-\angle{BAE}=180^{\circ}-\angle{EAC}$

meaning $AEPC$ is cyclic. it's percievable from this that $\angle{APC}=90^{\circ}$ .

Since $\angle{ACE}=\angle{BAE}$ , in other words, $ZA$ tangents circumircle of $\triangle{AEC}$ and $AB=BZ$ , $AZ$ is tangent to circumcircle of $\triangle{CEZ}$ . Also obtainable fact that, $\angle{AEZ}=180^{\circ}-\angle{ACZ}=\angle{ZFA}=\angle{AZT}=\angle{ZAT}$ . remembering $AB=BZ$ , combine with the recent fact $\triangle{ATZ}$ is isosceles, we have $\angle{ZBT}=90^{\circ}$ .

new notes: $TZBP$ is a cyclic quadrilateral.

Now it's left to show $BE$ is the radical axis of circum o' $TZBP$ and $ABE$ . using a previous highlight, $\angle{APC}=\angle{ZAC}=90^{\circ}$ , $=>$ $CP.CZ=CA^{2}$ . interestingly, they are the power of $C$ towards the 2 aforementioned circles respectivley. meaning $BC$ is the radical axis, and because $E$ lies on BC, we have also $BE$ the radical axis.

Since $E$ is on radical axis and it lies on circum o' $ABE$ , it also lies on circum o' $TZBP$ . hence $TZBE$ is cyclic too.

(Circum o' means Circumcircle of)

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1184 posts

L567

#14 h Mar 4, 2021, 7:06 AM

Y by

Since $DF$ is a diameter, $\angle DEP = 90$ . Observe that $\angle EPD = 90 - \angle EDP = 90 - \angle ZAE = 90 - \angle BAE = \angle ABE = \angle ABC = 90 - \angle ACB = \angle EAC$ , which means $EPCA$ is cyclic. Observe that $E$ is the C-humpty point in $\triangle CAZ$ and so $\angle ECZ = \angle EZB$ , which means $\angle EAP = \angle ECP = \angle ECZ = \angle EZB = \angle EZA$ which means $PA$ , $PZ$ are both tangents to $(AEZ)$ . But since $B$ is the midpoint of $AZ$ , $TB \perp AZ$ .

Now, to finish, observe that $\angle TZE = \angle TZA - \angle EZA = \angle TAB - \angle EAT = \angle BAE = 90 - \angle ABE = \angle TBE$ , which means $TEBZ$ is cyclic.

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#16 h Apr 4, 2021, 1:52 AM

Y by

Nice problem:

Key observation

Last steps

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#17 h Jun 15, 2021, 10:13 PM • 4 Y

Y by geometrylover123, Mango247, Mango247, Mango247

How about a solution involving inversion?

Let's apply inversion at $Z$ . We will drop the primes as they will be put to other use and denote $k(XYZ)$ to be the circumcircle of triangle $XYZ$ . Our goal becomes to prove that the points $B$ , $E$ and $T$ are collinear.

We get that $A$ is the midpoint of $ZB$ with $\measuredangle ZCA=90^{o}$ . Now, $E$ is the point on $k(ZBC)$ such that $\measuredangle AEZ=\measuredangle ACB$ . This means that $E$ is symmetric to $C$ with respect to the perpendicular bisector of $ZB$ and thus $\measuredangle AEB=90^{o}$ . $D$ is then the intersection of $AE$ and $ZC$ and $F$ is the point on $AE$ such that $FZD=90^{o}$ . Let $P'$ be the intersection of the perpendicular bisector of $ZB$ with $ZC$ . Since $Z$ , $C$ and $P'$ are collinear, so are $B$ , $E$ and $P'$ and thus $\measuredangle P'ED=90^{o}$ . This means that $ZFEP'$ is cyclic. This means that $P'$ is the intersection of $k(FEZ)$ with $ZC$ and thus $P\equiv P'$ . Note that $\measuredangle PAZ=90^{o}$ . Now let $T'$ be the intersection of $BE$ and a line through $Z$ parallel to $AE$ . Since $AE\parallel ZT'$ it follows that $\measuredangle PT'Z=90^{o}$ and thus $ZAPT'$ is cyclic and thus $T'$ is the intersection of $k(ZAP)$ with a line through $Z$ parallel to $AE$ (in the original image the tangent of $k(AEZ)$ at $Z$ ). Thus $T'\equiv T$ and from the definition of $T'$ it follows that $B$ , $E$ and $T$ are collinear, which completes the proof.

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322 posts

BVKRB-

#18 h Jul 16, 2021, 10:44 AM

Y by

Storage
Remarks
Diagram for reference

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Reason: Latex error

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#19 h Nov 3, 2021, 8:35 AM

Y by

storaij

btw, anyone tried showing $TA$ is tangent to $\odot AEZ$ :c , i was too high to complete this way thus left this way and for the above mentioned way :c

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#20 h Dec 19, 2021, 2:51 PM • 1 Y

Y by Math99989

∠EPD = 90 - ∠EDP = 90 - ∠EAB = EBA ---> PZBE is cyclic.
∠EFZ = ∠ABE = ∠CAE ---> AEPC is cyclic.
∠ZTP = 90 - ∠PZT = 90 - ∠DFZ = ∠ZDF = ∠ZEF ---> ZTPE is cyclic.
we had PZBE is cyclic and we now have ZTPE is cyclic so ZTPEB is cyclic.
we're Done.

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1162 posts

#21 h Aug 16, 2022, 12:28 AM • 1 Y

Y by Math99989

It's clear that $BZ^2 = BA^2 = BC \cdot BE$ so $E$ is the $C$ -Humpty point of $ACZ$ . Thus, $(ABC)$ and $(c)$ are symmetric about $BC$ .

Now, Thales' implies $\measuredangle AFZ = - \measuredangle ACZ = \measuredangle ZCA = 90^{\circ} - \measuredangle AZC = \measuredangle FZA$ which means $AF = AZ$ . Hence, $\measuredangle BEP = \measuredangle BEF = 90^{\circ} - \measuredangle FEA = 90^{\circ} - \measuredangle FZA = \measuredangle AZD = \measuredangle BZP$ so $BEPZ$ is cyclic. This yields $CP \cdot CZ = |Pow_{(BEPZ)}(C)| = CB \cdot CE = CA^2$ implying $AP \perp \overline{CPZ}$ . Now, because $B$ is the circumcenter of $(APZ)$ , $\measuredangle BPT = \measuredangle BPA = \measuredangle PAB = 90^{\circ} - \measuredangle AZP = \measuredangle FZA = \measuredangle AFZ = \measuredangle AZT = \measuredangle BZT$ so $BPTZ$ is cyclic. Thus, $BEPTZ$ is a cyclic pentagon, as required. $\blacksquare$

Remark: This problem only requires basic angle chasing and configurational knowledge.

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672 posts

#22 h Apr 8, 2023, 6:43 PM • 1 Y

Y by erkosfobiladol

$|BZ|^2=|BA|^2=|BE|.|BC| \implies \angle BCZ= \angle EZA=\angle EFA$
$\angle EZP+\angle EPZ=\angle FEZ =\angle FAZ=\angle DAE + \angle EAC=$
$\angle EZP+\angle EAC \implies \angle EPZ=\angle EAC$
$\implies A,E,P,C$ are cyclic and $AP \perp CZ$ and $|AB|=|BZ|=|BP|$ .
$|BP|^2=|BE|.|BC|$
$\angle BPE=\angle BCP\implies B,E,Z,P$ are cyclic.
$\angle EZT=\angle ZFE=\angle ZAE=\angle ACB=\angle APE$
$\implies P,E,Z,T$ are cyclic.
So $T,E,B,Z,P$ are cyclic.

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#26 h Feb 16, 2024, 9:46 PM

Y by

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.468690656675278cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.216542062234529, xmax = 20.455184579453665, ymin = -7.2798456425414235, ymax = 5.788863532528821; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); /* draw figures */ draw(circle((-1.4572411015647868,-2.3684620751896284), 3.1876612926324155), linewidth(0.8) + sexdts); draw((-4.295793764331847,-3.818911418408824)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((1.3455177968704262,-3.8869241503792575)--(-4.26,-0.85), linewidth(0.8) + wrwrwr); draw((-1.7925677789125065,0.8015127726610451)--(-4.295793764331847,-3.818911418408824), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-1.3747277729810385,4.475600514372738)--(-4.26,-0.85), linewidth(0.8) + wrwrwr); draw((-1.3747277729810385,4.475600514372738)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-1.7925677789125065,0.8015127726610451)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.26,-0.85)--(-2.7605089441992847,0.5406060654529699), linewidth(0.8) + wrwrwr); draw((-3.0083870072903296,1.4602120955349345)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-4.26,-0.85), linewidth(0.8) + dbwrru); draw((-4.26,-0.85)--(-4.295793764331847,-3.818911418408824), linewidth(0.8) + wrwrwr); draw((-4.26,-0.85)--(-1.439344219398863,-0.8840063659852172), linewidth(0.8) + dbwrru); draw((-1.439344219398863,-0.8840063659852172)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-1.439344219398863,-0.8840063659852172), linewidth(0.8) + dbwrru); draw((-1.439344219398863,-0.8840063659852172)--(1.3455177968704262,-3.8869241503792575), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-4.26,-0.85),linewidth(3.pt) + dotstyle); label("$A$", (-5.133764065799716,-1.2845043856712823), NE * labelscalefactor); dot((-1.439344219398863,-0.8840063659852172),linewidth(3.pt) + dotstyle); label("$B$", (-1.2944761315894897,-0.6603836870065672), NE * labelscalefactor); dot((-4.224206235668152,2.1189114184088234),linewidth(3.pt) + dotstyle); label("$C$", (-4.698770851578852,1.855011856096678), NE * labelscalefactor); dot((-2.7605089441992847,0.5406060654529699),linewidth(3.pt) + dotstyle); label("$E$", (-2.6751067680296203,0.7202469494335598), NE * labelscalefactor); dot((1.3813115612022742,-0.9180127319704345),linewidth(3.pt) + dotstyle); label("$Z$", (1.4478723928463864,-0.8116856745616496), NE * labelscalefactor); dot((-1.7925677789125065,0.8015127726610451),linewidth(3.pt) + dotstyle); label("$D$", (-1.7105565973659673,0.9093744338774129), NE * labelscalefactor); dot((-1.4572411015647868,-2.3684620751896284),linewidth(3.pt) + dotstyle); label("$O$", (-1.2944761315894897,-2.9677389972215744), NE * labelscalefactor); dot((-1.121914424217067,-5.538436923040301),linewidth(3.pt) + dotstyle); label("$F$", (-0.8216574204798558,-6.107255238989534), NE * labelscalefactor); dot((-3.0083870072903296,1.4602120955349345),linewidth(3.pt) + dotstyle); label("$P$", (-2.6940195164740057,1.5145823840977426), NE * labelscalefactor); dot((-1.3747277729810385,4.475600514372738),linewidth(3.pt) + dotstyle); label("$T$", (-1.04861040181248,4.200192663200456), NE * labelscalefactor); dot((1.3455177968704262,-3.8869241503792575),linewidth(3.pt) + dotstyle); label("$Q$", (1.6937381226233958,-4.121416652329078), NE * labelscalefactor); dot((-4.295793764331847,-3.818911418408824),linewidth(3.pt) + dotstyle); label("$R$", (-4.755509096912009,-4.140329400773463), NE * labelscalefactor); dot((-3.422880555013015,0.1410091949377491),linewidth(4.pt) + dotstyle); label("$S$", (-3.3937912089162636,0.3609047289902391), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Main Claim: $P$ is the midpoint of $CD$ .
Proof: Let $O$ be the center of $(c)$ and let $Q,R$ be the diametrically opposite point of $A,Z$ respectively. Then it is clear that $C,B,Q$ are collinear, and $C,A,R$ are collinear.
Now $\angle ADC = C = \angle ACD$ therefore, $AC = AD$ . Thus, $A$ is the circumcenter of the right triangle $\displaystyle \triangle CDR$ , and therefore $AD = AR$ . Thus, $(A,Q; R,D) = -1$ .
Thus, we have ( $S$ is the second intersection of $CF$ with $(c)$ )
$\begin{align*} (A,Q; R,D) = -1 &\xRightarrow{\text{take antipodes}} (Q,A; Z, F) = -1 \xRightarrow{\text{Invert at } C \text{ preserving } (c)} (E,R;D,S) = -1 \\ &\xRightarrow{\text{Project from }F \text{ to } CD} (P, \infty_{CD}; D, C) = -1 \end{align*}$ Thus, $P$ is the midpoint of $CD$ . $\spadesuit$

Now, it suffices to note that $AC = AD$ implies that $AP \perp CD$ and $AP$ is also the tangent to $(c)$ at $A$ . Therefore, $TA$ and $TZ$ are the two tangents to $(c)$ from $T$ , which implies that $\angle TBZ = 90^\circ$ . Also, on inverting at $C$ while preserving $(c)$ , we get that $(A,Q; R,D) = -1$ implies that $(R,E; A,Z) = -1$ , which implies that $T,R,E$ are collinear, and hence $\angle TEZ = 90^\circ$ . Thus, $T,E,B,Z$ are concyclic. (and one can see $P$ also is concyclic with them) $\square$

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#27 h Jun 5, 2024, 8:47 PM

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Rijul saini wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.468690656675278cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.216542062234529, xmax = 20.455184579453665, ymin = -7.2798456425414235, ymax = 5.788863532528821; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); /* draw figures */ draw(circle((-1.4572411015647868,-2.3684620751896284), 3.1876612926324155), linewidth(0.8) + sexdts); draw((-4.295793764331847,-3.818911418408824)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((1.3455177968704262,-3.8869241503792575)--(-4.26,-0.85), linewidth(0.8) + wrwrwr); draw((-1.7925677789125065,0.8015127726610451)--(-4.295793764331847,-3.818911418408824), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-1.3747277729810385,4.475600514372738)--(-4.26,-0.85), linewidth(0.8) + wrwrwr); draw((-1.3747277729810385,4.475600514372738)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-1.7925677789125065,0.8015127726610451)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.26,-0.85)--(-2.7605089441992847,0.5406060654529699), linewidth(0.8) + wrwrwr); draw((-3.0083870072903296,1.4602120955349345)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-1.121914424217067,-5.538436923040301), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-4.26,-0.85), linewidth(0.8) + dbwrru); draw((-4.26,-0.85)--(-4.295793764331847,-3.818911418408824), linewidth(0.8) + wrwrwr); draw((-4.26,-0.85)--(-1.439344219398863,-0.8840063659852172), linewidth(0.8) + dbwrru); draw((-1.439344219398863,-0.8840063659852172)--(1.3813115612022742,-0.9180127319704345), linewidth(0.8) + wrwrwr); draw((-4.224206235668152,2.1189114184088234)--(-1.439344219398863,-0.8840063659852172), linewidth(0.8) + dbwrru); draw((-1.439344219398863,-0.8840063659852172)--(1.3455177968704262,-3.8869241503792575), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-4.26,-0.85),linewidth(3.pt) + dotstyle); label("$A$", (-5.133764065799716,-1.2845043856712823), NE * labelscalefactor); dot((-1.439344219398863,-0.8840063659852172),linewidth(3.pt) + dotstyle); label("$B$", (-1.2944761315894897,-0.6603836870065672), NE * labelscalefactor); dot((-4.224206235668152,2.1189114184088234),linewidth(3.pt) + dotstyle); label("$C$", (-4.698770851578852,1.855011856096678), NE * labelscalefactor); dot((-2.7605089441992847,0.5406060654529699),linewidth(3.pt) + dotstyle); label("$E$", (-2.6751067680296203,0.7202469494335598), NE * labelscalefactor); dot((1.3813115612022742,-0.9180127319704345),linewidth(3.pt) + dotstyle); label("$Z$", (1.4478723928463864,-0.8116856745616496), NE * labelscalefactor); dot((-1.7925677789125065,0.8015127726610451),linewidth(3.pt) + dotstyle); label("$D$", (-1.7105565973659673,0.9093744338774129), NE * labelscalefactor); dot((-1.4572411015647868,-2.3684620751896284),linewidth(3.pt) + dotstyle); label("$O$", (-1.2944761315894897,-2.9677389972215744), NE * labelscalefactor); dot((-1.121914424217067,-5.538436923040301),linewidth(3.pt) + dotstyle); label("$F$", (-0.8216574204798558,-6.107255238989534), NE * labelscalefactor); dot((-3.0083870072903296,1.4602120955349345),linewidth(3.pt) + dotstyle); label("$P$", (-2.6940195164740057,1.5145823840977426), NE * labelscalefactor); dot((-1.3747277729810385,4.475600514372738),linewidth(3.pt) + dotstyle); label("$T$", (-1.04861040181248,4.200192663200456), NE * labelscalefactor); dot((1.3455177968704262,-3.8869241503792575),linewidth(3.pt) + dotstyle); label("$Q$", (1.6937381226233958,-4.121416652329078), NE * labelscalefactor); dot((-4.295793764331847,-3.818911418408824),linewidth(3.pt) + dotstyle); label("$R$", (-4.755509096912009,-4.140329400773463), NE * labelscalefactor); dot((-3.422880555013015,0.1410091949377491),linewidth(4.pt) + dotstyle); label("$S$", (-3.3937912089162636,0.3609047289902391), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Main Claim: $P$ is the midpoint of $CD$ .
Proof: Let $O$ be the center of $(c)$ and let $Q,R$ be the diametrically opposite point of $A,Z$ respectively. Then it is clear that $C,B,Q$ are collinear, and $C,A,R$ are collinear.
Now $\angle ADC = C = \angle ACD$ therefore, $AC = AD$ .

Why $\angle ADC = C = \angle ACD$ ?

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