Is there any typo error in this problem?

Let $N$ be the midpoint of $DE$. The condition implies $CN=r_A \Leftrightarrow \sin{\frac{\gamma}{2}} = \frac{s-b}{s-c}$ (by fiddling around with $P=r_A \cdot (s-a)$ and $P=\sqrt{s(s-a)(s-b)(s-c)}$ and $\sin{\frac{\gamma}{2}} = \sqrt{\frac{(s-a)(s-b)}{ab}}$) then $DE=DX$ and $DZ=DY$ where $Y$ is the touchpoint of the $A-$excircle with $BC$. Then $XZYE$ is cyclic. Notice that the circumcenter $S$ of triangle $EDX$ lies on the incircle and is in fact the second intersection of $AI$ with the incircle since $AI$ bisects $DE$. $S$ is the midarc. Lemma: Let $\omega_1,\omega_2$ be two circles. Let $A$ be the intersection of the exterior common tangents. A line $p$ cuts $\omega_1$ at $U,V$ and $\omega_2$ at $W,Z$. The midarcs $UV,WZ$ are collinear with $A$. (in general any parallelish lines could work here) Proof: Add the circles' centers and use the ratio from the homothety. From this we have $A,S,Z$ are collinear (degenerate case for tangency, apply to $p=DE$). Now focus on the isosceles trapezoid $XZYT$. The problem is equivalent to the following: Given isosceles trapezoid $EXZY$ and $D$ the intersection of its diagonals. Let $S$ be the circumcenter of $EDX$ and $K=ESD \cap EXYZ$. $B$ is the midpoint of $DX$ then $BK$ and $EZ$ concur on $EDSK$. Proof: Angles. Add the midpoint of $YZ$ say $G$. By angles get $KSBGY$ cyclic. Notice that it's enough $\angle ZSD = \angle DKB$ but $\angle ZSD = \angle YSD = \angle YSG = \angle YKG$ but also $\angle KGY = \angle DBK$ so I just need $\angle KYG = \angle KDB$ which is true because they are both $180^{\circ} - \angle KEZ$.

Iranian TST 2022/3 wrote: Incircle $\omega$ of triangle $ABC$ is tangent to sides $CB$ and $CA$ at $D$ and $E$, respectively. Point $X$ is the reflection of $D$ with respect to $B$. Suppose that the line $DE$ is tangent to the $A$-excircle at $Z$. Let the circumcircle of triangle $XZE$ intersect $\omega$ for the second time at $K$. Prove that the intersection of $BK$ and $AZ$ lies on $\omega$. Nice configuration and problem. Let $F$ be the $A$-excircle touchpoint with line $BC$. Claim 01. $DX = DE$ and $XEFZ$ is cyclic. Proof. Let $EC$ be tangent to $A$-excircle at $G$. Now, as $DE$ is tangent to $A$-excircle at $Z$. By the condition, we have $DF = DZ$ and $EG = EZ$. However, \begin{align*} EZ &= EG = EC + CG = DC + CF = DF + 2FC = DF + 2BD = DZ + XD \implies ED = XD \end{align*}Therefore, we have $XZFE$ is cyclic as well as $DX = DE$ and $DF = DZ$. Let $W$ be map of $Z$ under the homothety $\Phi$ which sends $A$-excircle to incircle from $A$. Claim 02. $W$ is the circumcenter of $\triangle XDE$. Proof. We will first prove that $WD = WE$. To prove this, it suffices to prove that $WI \perp DE$. By homothety $\Phi$, we have $WI \parallel ZI_A$. Therefore, we just need to prove that $DE \perp ZI_A$, however this is clearly true as $DE$ is tangent to $A$-excircle at $Z$. Now we will prove that $WX = WE$. To prove this, note that $\measuredangle BDW = \measuredangle DEW = \measuredangle WDE$, i.e. $WD$ bisects $\angle BDE$. Similarly, $DI_A$ bisects $\angle BDE$ as well. Therefore, $W,D,I_A$ are collinear. However, $D$ and $I_A$ lies on the perpendicular bisector of $FZ$, which coincides with the perpendicular bisector of $EX$, then we have $W$ lies in the perpendicular bisector of $EX$, i.e. $WX = WE$. Let $V$ be the midpoint of $FZ$. Note that we have $W,D,V$ being collinear as they all lie on the perpendicular bisector of $FZ$. Moreover, \[ \measuredangle WKF = \measuredangle WKE + \measuredangle EKF = \measuredangle WDE + \measuredangle EZF = \measuredangle VDZ + \measuredangle DZV = 90^{\circ} = \measuredangle WVF = \measuredangle WBF \implies WVFBK \ \text{cyclic} \]Now, let us suppose that $WZ$ intersects the incircle at $Q$. We claim that $B,K,Q$ are collinear, which finishes the problem. Indeed, we have $\measuredangle QWD = \measuredangle ZWD = \measuredangle DWF$, and $\measuredangle FDW = \measuredangle DQW$, which implies $\measuredangle WDQ = \measuredangle WFD = \measuredangle WVB$, i.e. $DQ \parallel BV$. However, this implies \[ \measuredangle(DQ, QK) = \measuredangle DQK = \measuredangle DEK = \measuredangle ZEK = \measuredangle ZFK = \measuredangle VFK = \measuredangle VBK = \measuredangle (VB, BK) \stackrel{BV \parallel DQ}{=} \measuredangle (DQ,BK) \]which implies $B,K,Q$ being collinear.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -12.54, xmax = 11.56, ymin = -10.14, ymax = 6.08; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-7.894,4.691)--(-7.77,-1.15)); draw((-7.776021225931557,-1.1528169832738713)--(1.2237085798166198,-1.3249469288874225)); draw((-7.894317653149731,4.691532756425198)--(1.2237085798166198,-1.3249469288874225)); draw(circle((-5.775965991872319,0.8494687674002902), 2.0401659334952345)); draw(circle((-0.8667587100054215,-8.054382683281778), 6.768180474240545)); draw((-7.894317653149731,4.69153275642529)--(xmax, -49.404279377942295*xmax-385.3215420780058), dotted); /* ray */ draw((-7.894317653149731,4.691532756425198)--(xmax, -0.659844524636259*xmax-0.5174895227450136), dotted); /* ray */ draw((-4.6523409102629385,2.5523321536332557)--(xmin, 3.2191059803510376*xmin + 17.52871060049247), ccqqqq); /* ray */ draw(circle((-5.393588570644753,-1.7785050676965508), 4.393813734294753), qqwuqq); draw((-7.776021225931557,-1.1528169832738713)--(-7.338269751430542,2.161522119617616), linetype("2 2") + blue); draw((-7.894317653149731,4.691532756425198)--(-7.2367169670277525,-5.767048266174389), linetype("4 4") + blue); draw(circle((-7.696093843549012,1.5389512323041699), 3.314745201696238), dotted); /* dots and labels */ dot((-7.894317653149731,4.691532756425198),dotstyle); label("$A$", (-7.79225017858806,4.926247138170736), NE * labelscalefactor); dot((-7.776021225931557,-1.1528169832738713),dotstyle); label("$B$", (-7.676335171577308,-0.9158692151712193), NE * labelscalefactor); dot((1.2237085798166198,-1.3249469288874225),dotstyle); label("$C$", (1.3186693724571115,-1.1013332263884243), NE * labelscalefactor); dot((-5.775965991872319,0.8494687674002902),dotstyle); label("$I$", (-5.682597050992359,1.0315029026094324), NE * labelscalefactor); dot((-0.8667587100054215,-8.054382683281778), dotstyle); label("$I_A$", (-0.7678007537364392,-7.870769635816404), NE * labelscalefactor); dot((-5.814979322849821,-1.1903241133110145),dotstyle); label("$D$", (-5.72896305379666,-1.0086012207798218), NE * labelscalefactor); dot((-4.6523409102629385,2.5523321536332553), dotstyle); label("$E$", (-4.569812983689132,2.747045006368578), NE * labelscalefactor); dot((-0.7373333232651172,-1.2874397988502788), dotstyle); label("$F$", (-0.6518857467256864,-1.1013332263884243), NE * labelscalefactor); dot((-9.737063129013293,-1.1153098532367283), dotstyle); label("$X$", (-9.64,-0.96), NE * labelscalefactor); dot((-7.2367169670277525,-5.767048266174389), dotstyle); label("$Z$", (-7.143126139327845,-5.575652497003493), NE * labelscalefactor); dot((-7.338269751430542,2.161522119617616), dotstyle); label("$K$", (-7.4,2.3529339825320177), NE * labelscalefactor); dot((-7.696093843549012,1.5389512323041699), dotstyle); label("$W$", (-7.6,1.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

A wise man once said, that Reim's shall save ur day, 45 mins solve on a friends call with ACBY, really fun problem 10/10. Let the $A$-excircle be tangent to $BC,AC$ at $L,G$ respectivily, $AZ \cap \omega=O,P$ where $A,O,P,Z$ are colinear in this order, $CI \cap DE=C'$ where $I$ is the incenter of $\triangle ABC$ and let $I_A$ be the $A$-excenter, let $Q$ the midpoint of $ZL$. Claim 1: $XELZ$ is a ISLsosceles trapezoid. Proof: Note $XL=BC=BD+DC=CG+CE=EG=EZ$ and also $DZ=DL$ so $XD=DE$ hence by PoP and the equal sides we get our desired claim. Claim 2: $O$ is the center of $(XDE)$ Proof: By homothety and tangency we get $OI \parallel ZI_A \parallel CI$ so $O,I,C$ are colinear hence $OD=OE$, by ratios: $$\frac{OI}{IC}=\frac{ID}{IC}=\frac{C'D}{CD}=\frac{BD}{CD} \implies OB \parallel ID \perp BC \implies OX=OD \implies O \; \text{center of} \; (XDE)$$Claim 3: $BOKLQ$ is cyclic. Proof: First clearly $\angle OQL=90=\angle OBL$ so $OBQL$ is cyclic, now $\angle KLQ=\angle KEZ=180-\angle KOQ$ so $OKLQ$ is cyclic hence joining both results we get our desired claim. Finishing: $\angle BDP=\angle POD=\angle LOQ=\angle LBQ$ so $PD \parallel BQ$ but by Reim's this implies that $B,P,K$ are colinear so $BK,AZ$ meet at $P$ which lies in $\omega$ as desired thus we are done .

Very nice and very hard Let $I$ and $I_A$ be the incenter and $A$-excenter, let the $A$-excircle touch $BC$ at $F$, let $CI$ and $DI$ meet $w$ past $I$ at $G$ and $H$, and let $BK$ meet $w$ again at $P$. Claim: The points $A$, $G$, and $Z$ are collinear Consider the homothety centered at $A$ mapping the incircle to the excircle. This homothety takes $H$ to $F$ and we claim it also takes $G$ to $Z$, but this follows from (notice we preserve direction throughout) $$\angle HIG=\angle DIC=\angle DEC=\angle CDE=\angle FI_A Z$$ Claim: $DE=DX$ By Pitot's Theorem for tangential quadrilaterals we have that $$AE+DE=AB+BD$$Notice as $AB=AE+BD$ we have that $DE=2BD=DX$ (this also gives that $EFXZ$ is an isosceles trapezoid). Now let $\zeta$ be the negative inversion centered at $D$ that preserves $(EFXZ)$. By construction, $\zeta\colon E\leftrightarrow Z$ and $\zeta\colon F\leftrightarrow X$. Notice that $\zeta$ maps $w$ to the line through $Z$ parallel to $BC$. Let $w$ map $B$, $G$, $K$, and $P$ to $B'$, $G'$, $K'$, and $P'$. As $FK'Z'X$ and $DB'P'K'$ are both cyclic they both must be isosceles trapezoids. Let $K_1$ be the foot of $K$ onto $BC$. Then $$P'Z=P'K'+K'Z=(B'D-2K_1D)+(FX-2FK_1)=B'D'+FX-2FD=BF+DX=FX$$Then as $GE=GD$ we must have $ZD=ZG'$. As $ZP'=XF=ZE$ we have that $EDG'P'$ is a cyclic isosceles trapezoid. But this means that $G$, $Z$, and $P$, are collinear, so we are done.