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
Problem
Source: JBMO 2020
Tags: Junior, geometry, cyclic quadrilateral, Balkan
i3435
11.09.2020 16:55
Claim: $P,E,B,Z$ and $C,A,E,P$ are cyclic.
Proof:$\measuredangle DEP= \measuredangle DEF=90$, so $\measuredangle EPZ=\measuredangle EPD=90-\measuredangle PDE=90-\measuredangle ZDE=90-\measuredangle ZAE=\measuredangle EBA=\measuredangle EBZ$. Thus $P,E,B,Z$ are cyclic. $\measuredangle CPE=\measuredangle ZPE=\measuredangle ZBE=\measuredangle EBA=\measuredangle CAE$, so $C,A,E,P$ is cyclic.
Let $T'$ be the intersection of $PA$ and $(PEBZ)$. Since $\measuredangle CPA=\measuredangle CEA=90$, $\measuredangle ZPT'=\measuredangle CPA=90$. Then $\measuredangle T'ZB=90-\measuredangle BT'Z=90-\measuredangle BEZ=90+\measuredangle ZEB=\measuredangle ZEA$, so $T'Z$ is tangent to $(c)$.
Note that $AB=BZ$ is extraneous.
Steff9
11.09.2020 16:56
$\angle ABE= 90-\angle BAE\equiv 90-\angle ZAE=90-\angle ZFE\equiv 90 -\angle ZFP=\angle ZPF\equiv\angle ZPE$, so $ZPEB$ is cyclic.
From Euclid's Laws, we have $BE\cdot BC=BA^2=BZ^2$, so $\triangle BEZ\sim\triangle BZC$ and therefore $\angle BEZ=\angle BZC\equiv \angle BZP$.
Combining the above two results, we get $\overarc{BZ}=\overarc{BP}$, i.e. $BP=BZ=BA$, so $\triangle AZP$ is right and thus $\angle ZPT=90$. Let $\angle ZTP=\varphi$. Then, $\angle PZT=90-\varphi$. Let $O$ be the center of $(c)$. Since $\angle OZT=90=\angle FZD$, we have $\angle FZO=\angle PZT=90-\varphi$. Since $OF=OZ$, we get $\angle FOZ=2\varphi$ and therefore $\angle FEZ=\varphi$. Thus, $\angle ZTP=\varphi=\angle ZEF$, so $ZTPE$ is cyclic $\blacksquare$
dangerousliri
11.09.2020 20:29
This problem was proposed by Theoklitos Parayiou, Cyprus.
Steve12345
11.09.2020 21:41
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 )
Attachments:
VicKmath7
12.09.2020 10:55
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.
geometry6
12.09.2020 16:15
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.$\blacksquare$
Attachments:
Math00954
12.09.2020 16:55
Let $BC$ intersect $(c)$ at $M$. It's obvious that $M$ and $A$ are antipodes. Now, $OB$ is the $M$-midline in $\triangle AMC$ and the $A$-midline in $\triangle AMZ$, so $AM||ZC$. After this, angle-chase to prove that $AF=AZ$. Since $\angle FZC=90^o$ (let $P'$ be the intersection of $CZ$ and the $A$-tangent to $(c)$), we have that $AP'EC$ is cyclic. Angle-chase some more to prove $F-E-P'$. It's another easy angle-chase now to prove that $BZPE$ is cyclic, and since $T$ is the intersection of the two tangents, we have that $\angle TBZ= \angle TPZ=90^o$, so $TPBZ$ is cyclic. Finally, $TPEBZ$ is cyclic, and we are done.
coldheart361
13.09.2020 07:23
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)
L567
04.03.2021 10:06
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.
CROWmatician
04.04.2021 04:52
Nice problem:
The main idea is to prove that $|TZ|=|TA|$
Claim: $AEPC$ is cyclic
Proof: Note that $$\angle{FEA}=\angle{FZA}=\angle{FZD}-\angle{AZD}=90^o-\angle{AZC}=\angle{ZCA}$$Done.
Now let $Y$ be the second intersection of $\overline{AO}$ with $(c)$ other than $A$,
clearly $C, B,$ and $Y$ are collinear.
Claim: $YZCA$ is a parallelogram
Proof: See that $$\triangle{YZB}\cong \triangle{CAB}$$so $B$ is midpoint of $|YC|$, now in quadrillateral $YZCA$ diagonals bisects each other, therefore it's a parallelogram.
Done.
Claim: $PA$ is tangents to $(c)$ at $A$
Proof: We are going to use results from two previous claims, observe that $$\angle{EAP}=\angle{ECP}=\angle{YCZ}=\angle{AYC}=\angle{AYF}$$so $PA$ is tangent to circumcircle of $(c)$ at point $A$.
Done.
Last claim means that $TA$ is tangents to $(c)$ at $A$ as well, then using power of point $T$ with respect to $(c)$ we can easily get that $$|TZ|=|TA|$$as desired.
Using previously stated key idea we derive that $B$ is midpoint of base in isosceles triangle, so $\angle{ZBT}=90^o$, we do some angle chasing to finish
$$\angle{ZTB}=90^o-\angle{BZT}=\angle{OZB}=\angle{OZA}=\angle{OAZ}=\angle{YAZ}=\angle{YEZ}=\angle{BEZ}$$so $TEBZ$ is cyclic, and we are done. $\blacksquare$
NZP_IMOCOMP4
16.06.2021 01:13
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.
BVKRB-
16.07.2021 13:44
Claim: $AEPC$ is cyclic
Proof$\angle FED = \angle FZD = \angle DEP = 90^{\circ} $ and $$\angle BCA = \angle BAE = \angle EFZ \implies \angle FPZ = 90^{\circ} - \angle BCA = \angle EAC$$Therefore $AEPC$ is cyclic as desired $\blacksquare$
Now by the above statement we have $$\angle ZFE = \angle ECA=\angle EPA = \angle FPA$$Thus $ZF \parallel AP \equiv AT$
Claim: $ZBEP$ is cyclic
Proof$\angle ZBE = \angle ZBC = 180^{\circ} - \angle ABC = 180^{\circ} - \angle EAC = 180^{\circ} - \angle ZPE$ Therefore $ZBEP$ is cyclic
Now by the above claim's we get that $$\angle ZEP = \angle ZED + \angle DEP = \angle ZAD + \angle DEP = \angle ZAD + 90^{\circ} = \angle ZAE - \angle DAE +90^{\circ} $$$$= \angle EZT - \angle EZD + \angle FZD = (\angle FZD - \angle EZD) + \angle EZT = \angle FZT = 180^{\circ} - \angle ZTP$$And so $ZEPT$ is cyclic $\implies $ $ZBEPT$ is cyclic and so we are done! $\blacksquare$
This was an absolutely amazing problem! Though there is no use of the fact that $AB=BZ$ one can easily notice after solving the problem that $\angle EBT = \angle EZT = \angle BAE = 90^{\circ} - \angle ABE$ so we get that $\angle ABT=90^{\circ} \implies ABZ$ is isosceles!
Diagram for reference
Attachments:
REYNA_MAIN
03.11.2021 11:35
$\angle DEP = 90^{\circ}$, $\measuredangle EAZ = \measuredangle EDZ = \measuredangle EDP \implies \triangle BEA \sim \triangle PED \implies \measuredangle EPC = \measuredangle EPD = \measuredangle EPZ = \measuredangle EBA = \measuredangle EBZ = \measuredangle EAC \implies AEPC$ and $BEPZ$ are cyclic.
Now $\angle TZP = \angle ZAD = 90^{\circ} - \angle ZAF = 90^{\circ} - \angle ZEF = \angle ZEP - 90^{\circ}$, $90^{\circ} = \angle AEC = \angle APC = \angle TPZ \implies \angle PTZ = 90^{\circ} - \angle TZP \implies \angle PTZ + \angle PEZ = 90^{\circ} - \angle TZP + 90^{\circ} + \angle TZP = 180^{\circ} \implies ZEPT$ is cyclic $\implies ZEBPT$ is cyclic and we are done .
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
Mahdi_Mashayekhi
19.12.2021 17:51
∠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.
ike.chen
16.08.2022 03:28
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.
bin_sherlo
08.04.2023 21:43
$|BZ|^2=|BA|^2=|BE|.|BC| \implies \angle BCZ= \angle EZA=\angle EFA$ $\angle EZP+\angle EPZ=\angle FEZ =\angle FAZ=\angle DAE + \angle EAC=$ $\angle EZP+\angle EAC \implies \angle EPZ=\angle EAC$ $\implies A,E,P,C$ 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.
Rijul saini
17.02.2024 00:46
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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) $\square$
PennyLane_31
05.06.2024 23:47
Rijul saini wrote: [asy][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][/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 $\angle ADC = C = \angle ACD$?