Too easy for #3. Let $Y$ be the intersection of the bisector of $\angle{CXB}$ with $BC$. Note that $\angle{IXA}=\angle{IAC}=\angle{BXY}$ and $\angle{XIA}=180^0-\angle{XAC}=\angle{XBY}$, so $\triangle{XBY}\sim\triangle{XIA}.$ Therefore, $X$ is the center of spiral similarity that sends $IA$ to $BY$, meaning that $Z=AY\cap BI$ is on $(XIA).$ From here it is just angle chasing: $\angle{ZAC}=180^0-\angle{ZIA}=90^0-\frac{\angle{C}}{2}$, meaning that $\angle{ZAD}=90^0-\frac{\angle{C}}{2}-\angle{B}=\frac{\angle{A}-\angle{B}}{2}.$ Done!

Easy . Here's my solution: Suppose the internal angle bisector of $\angle BAD$ meets $BC$ at $T$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$. Using the inversive distance formula, one easily gets the following equivalent problem:- Inverted problem wrote: Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I_A$ be its $A$-excenter. Denote the circumcircle of $\triangle ABC$ by $\Omega$, and let $D$ be a point on $\Omega$ such that $BD \parallel AC$. Suppose the line through $I_A$ parallel to $AC$ meets $BC$ again at $X$, and let $T$ be the midpoint of arc $\widehat{BD}$ not containing $A$. Prove that $TX$ bisects $\angle BTC$ externally. Let $M$ be the midpoint of arc $\widehat{BC}$ not containing $A$. Then we just need to show that $M,T,X$ are collinear. As $TB=TD$, and $ABDC$ is an isosceles trapezoid, so we get that $TA=TC$ is also true, which means that $B,T,I_A$ are collinear. By Pascal on $BTTMAC$, and using the fact that the tangent to $\Omega$ at $T$ is parallel to $AC$, we get the desired conclusion. $\blacksquare$ EDIT: @below Sniped

Perform $\sqrt{bc}$ inversion. After some simple angle chasing the problem now reads as follows: Quote: Let $ABC$ be a triangle with $A$-excenter $I_A$ and let $AI_A$ cut $BC$ at $M$. Let $X$ lie on $BC$ such that $I_AX \parallel AB$, and let the circumcircle of $AMX$ cut the circumcircle of $ABC$ again at $Y$. Show that $\angle CAY = \frac{\angle A - \angle B}{2}$. This is because if $N$ is the midpoint of minor arc $BC$ then the bisector of $\angle CXB$ is line $XN$, which maps to circle $AMX$ in the new diagram. Keeping $N$ as the midpoint of this arc, inversion at $N$ with radius $NB = NC$ gives that $N, X, Y$ are collinear. However by Reim's theorem $NCXI_A$ is cyclic, and then easy angle chasing shows that $XY$ bisects angle $MXZ$. Now $$\angle CAY = \angle CAM - \angle YAM = \frac{\angle A}{2} - \angle MXY = \frac{\angle A}{2} - \frac{\angle MXZ}{2} = \frac{\angle A - \angle B}{2}$$ Done.

Assassino9931 wrote: Extremely easy. It suffices to show $\frac{DA}{AB} = \frac{CX}{XB}$. Wait! How are you supposed to prove this? In fact if $T$ is the said concurrency point, then we have $\tfrac{DA}{AB}=\tfrac{DT}{BT}$ and $\tfrac{CX}{XB}=\tfrac{CT}{BT}$, which is definitely not same as what you have supposedly proven.

Somehow I found this problem easier than P1. Let the angle bisector of $\angle DAB$ meet $BC$ at $T$. Then by angle chasing, $CA = CT$. Now let $\omega$ intersect $AB$ again at $Y$. Then $Y$ is reflection across $A$ through intouch point thus $AY = b+c-a$ $\implies BY = a-b = BT$. Notice that $X$ is spiral center from $BY\to AC$. Thus $$\frac{BX}{XC} = \frac{BY}{AC} = \frac{BT}{TC}$$so we are done.

math_pi_rate wrote: Assassino9931 wrote: Extremely easy. It suffices to show $\frac{DA}{AB} = \frac{CX}{XB}$. Wait! How are you supposed to prove this? In fact if $T$ is the said concurrency point, then we have $\tfrac{DA}{AB}=\tfrac{DT}{BT}$ and $\tfrac{CX}{XB}=\tfrac{CT}{BT}$, which is definitely not same as what you have supposedly proven. My bad. Instead, if $AY$ is the bisector of $\angle DAB$ and $XT$ - of angle $CXB$, then we want $CT = CY$. But since $CT = CA$, we reduce to $\frac{CX.CB}{CX+XB} = CA$ and then the computation of $CX, XB$ as in my previous attempt, etc.

Here is a cross-ratio/trig solution: rare for me to do one of these, but this problem called out to me that way. As usual, let $\alpha = \angle BAC$, etc. Let $T$ be the foot of the bisector of $\angle BAD$ on $\overline{BD}$, so that \[ \frac{TB}{TC} = \frac{AB \sin \angle BAT}{AC \sin \angle CAT} = \frac{AB \sin \frac{\alpha-\beta}{2}} {AC \sin \frac{\alpha+\beta}{2}}. \]Also, call $(ABC)$ by $\Gamma$ and let ray $XI$ meet $\Gamma$ again at $W$, meaning that $\angle WXA = \angle IXA = \angle IAC = \frac{\beta}{2}$, since we assume $\overline{AC}$ was tangent to $(IAX)$. Thus arc $\widehat{AW}$ also has measure $\beta$. [asy][asy] pair A = dir(50); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair M_a = circumcenter(B, I, C); pair M_b = circumcenter(C, I, A); pair M_c = circumcenter(A, I, B); pair W = A*B/M_a; pair X = -W+2*foot(origin, I, W); pair D = extension(B, C, A, C*C/A); filldraw(unitcircle, invisible, lightblue); draw(A--B--C--cycle, lightblue); draw(X--W, lightred); draw(B--M_b, lightred); draw(C--M_c, lightred); draw(A--M_a, lightred); filldraw(W--M_b--A--M_c--cycle, invisible, deepgreen); filldraw(X--B--M_a--C--cycle, invisible, deepgreen); draw(A--D, lightblue); pair T = extension(A, incenter(A, B, D), B, D); draw(X--T--A, dashed+brown); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$M_a$", M_a, dir(M_a)); dot("$M_b$", M_b, dir(M_b)); dot("$M_c$", M_c, dir(M_c)); dot("$W$", W, dir(W)); dot("$X$", X, dir(X)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); /* TSQ Source: A = dir 50 B = dir 210 C = dir 330 I = incenter A B C M_a = circumcenter B I C M_b = circumcenter C I A M_c = circumcenter A I B W = A*B/M_a X = -W+2*foot origin I W D = extension B C A C*C/A unitcircle 0.1 lightcyan / lightblue A--B--C--cycle lightblue X--W lightred B--M_b lightred C--M_c lightred A--M_a lightred W--M_b--A--M_c--cycle 0.1 yellow / deepgreen X--B--M_a--C--cycle 0.1 yellow / deepgreen A--D lightblue T = extension A incenter A B D B D X--T--A dashed brown */ [/asy][/asy] Now, \begin{align*} \frac{XB}{XC} &= -(BC;XM_a)_\Gamma \overset{I}{=} -(M_b M_c; W A)_\Gamma \\ &= -\frac {\sin \frac{1}{2} \mathrm{m}\widehat{M_b W}} {\sin \frac{1}{2} \mathrm{m}\widehat{M_c W}} \div \frac {\sin \frac{1}{2} \mathrm{m}\widehat{M_b A}} {\sin \frac{1}{2} \mathrm{m}\widehat{M_c A}} = \frac{\sin\frac{\alpha-\beta}{2}} {\sin\frac{\alpha +\gamma}{2}} \div \frac {\sin \frac{\gamma}{2}} {\sin \frac{\beta}{2}} \\ \end{align*}Therefore, \[ \frac{XB}{XC} \div \frac{TB}{TC} = \frac{\sin\frac{\alpha+\beta}{2} \sin\frac{\gamma}{2}} {\sin\frac{\alpha+\gamma}{2} \sin\frac{\beta}{2}} \div \frac{AB}{AC} \\ = \frac{2\cos \frac{\gamma}{2} \sin \frac{\gamma}{2}} {2\cos \frac{\beta}{2} \sin \frac{\beta}{2}} \div \frac{AB}{AC} = 1 \]the last step being the law of sines $AB/AC = \sin\gamma/\sin\beta$.

Let $Y$ be the foot of the angle bisector of $\angle BXC$ on $BC$. Let $M$ be the midpoint of $\widehat{BC}$ not containing $A$ and finally let $F \equiv AI \cap BC$. Note $X,M,Y$ are colinear. Inversion in $\odot BIC$ (which has centre $M$) takes $\odot XAI \rightarrow \odot YFI$. By the tangency condition for $\omega$ this is tangent to $\odot FMC$. Hence $YI \parallel MC$. Then: $$\angle FYI=\angle MCB=\frac{\angle A}{2}=\angle BAI$$So $ABYI$ is cyclic. Then: $$\angle BAY=\angle BAI-\angle YAI=\angle BAI-\angle YBI=\frac{\angle A-\angle B}{2}=\frac{\angle BAD}{2}$$Hence the two angle bisectors concur at $Y$

This problem was proposed by Burii.

Let $AD$ intersect the circumcircle of $\triangle ABC$ second time at $E$. Then $C$ is the midpoint of the arc $AE$ not containing $X$. Draw the circle touching the circumcircle of $\triangle ABC$ at $X'$, segment $DB$ at $K$ and segment $AD$ at $J$. We prove that $X'=X$. First, by Sawayama's lemma $I\in KL$. Then $\angle IJA=90^{\circ}-\frac{\angle KDJ}{2}=90^{\circ}-\frac{180^{\circ}-\angle BAC}{2}=\frac{\angle BAC}{2}=\angle IAC$, hence the circumcircle of $AIJ$ is tangent to $AC$ at $A$. This means that it coincides with $\omega$ and intersects the circimcircle of $\triangle ABC$ at $X$. Then, by Archimede's lemma $XJ$ passes through $C$, so $\angle AX'C=\angle ABC=\angle DAC=180^{\circ}-\angle AIJ$, hence $AIJX'$ is concyclic, and indeed $X'=X$. It follows from Archimede's lemma that $XK$ is the bisector of $\angle BXC$, and since $BC$ is the bisector of $\angle ABE$, it's left to prove that $K$ is incenter of triangle $ABE$. But, again by Archimede's lemma $CE^2=CA^2=CJ\cdot CX=CK^2$, which means that $CE=CK=CK$, and by incenter lemma we get that $K$ is indeed incenter of triangle $ABE$, what was required to be proved.

Let \(K\) be the intersection of the angle bisector of \(\angle DAB\) and \(BC\). We claim \(BK=a-b\), where \(BC=a, AC=b, AB=c\). Proof is as follows. By similar triangles we have \(DC=b^2/a\), so \(BD=(a^2-b^2)/a\). Now by stewart's theorem, we can compute \(AD=d\), as \[\dfrac{b^2c^2}{a}+\frac{b^2(a^2-b^2)}{a}=ad^2+\frac{a(a^2-b^2)b^2}{a^2}.\] Which reduces to \(b^2c^2=a^2d^2\) (the second terms on each side cancel out), so \(d=bc/a\). Finally, by angle bisector theorem, we have \(BK/BD=a/(a+b)\), so \(BK=a-b\), as desired (note this makes sense because the angle condition yields \(a>b\)). Now we use barycentric coordinates. First, we find line \(KM\), where \(M\) is the midpoint of arc \(BC\) not containing \(A\). We claim \(X\) lies on \(KM\). This finishes, as it implies that the angle bisector of \(\angle BXC\) lies on \(K\). We can compute \(K:(0:b:a-b)\), \(M:(-a^2/(b+c):b:c)\). We can then compute line \(KM\) to be \[\dfrac{b(b+c-a)(b+c)}{a^2}x+(b-a)y+bz=0.\] Now compute the desired circle. Interpret the tangency condition as \((m:0:n)\) lies on the circle if and only if \(n=0\). This gives us \(a^2yz+b^2xz+c^2xy=(x+y+z)(vy+wz)\) where \(vb+wc=abc\). The tangency condition gives \(b^2mn=(m+n)(wn)\). We basically want a \(w\) such that if \(n\neq 0\) then this has no solutions. Assume \(n\neq 0\), so \(b^2m=(m+n)w\). Moreover, the point at infinity doesn't matter as a point of intersection, so \(m+n\neq 0\). This means that when \(m+n\neq 0\) and \(n\neq 0\), we can see \(w=\frac{b^2m}{(m+n)}\) has no solutions \(m,n\) upon fixing \(w\). if \(w=b^2\) this is true, but if \(w\neq b^2\) then we can always find a pair \(m,n\) that works, contradiction. Thus, \(w=b^2\) (not sure if this step is correct), and the circle's equation is \[a^2yz+b^2xz+c^2xy=(x+y+z)((ac-bc)y+b^2z)\]. Thus, point \(X:(x,y,z)\) lies on the circumcircle \(a^2yz+b^2xz+c^2xy=0\) and the above circle, so \(0=(x+y+z)((ac-bc)y+b^2z)\). Clearly \(x+y+z\neq 0\) so \((ac-bc)y+b^2z=0\). This gives \(X\) to be of the form \((t:b^2:bc-ac)\). Plugging this into the circumcircle equation gives \[X:\left(\dfrac{a^2(a-b)}{b+c-a}:b^2:bc-ac\right).\] Plugging these coordinates into line \(KM\) finishes.

juckter wrote: Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$. Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$. Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$ intersect at a point on line $BC$. Let $Y$ be the intersection of the angle bisector of $\angle CXB$ with $BC$ and $M$ be the midpoint of arc $BC$ not containing $A$. As $\omega$ is tangent to $AC$, $\angle AXI = \angle CAI = \angle CAM = \angle CBM$. Further $M,Y,X$ are collinear so $\angle AXM = \angle ABM = \angle ABY + \angle CBA$ giving $\angle IXY = \angle B$. From shooting lemma, $MI^2 = MX\cdot MY$ giving $\angle MIY = \angle IXM = \angle B$ so $A,B,Y,I$ are concyclic. This gives \[\angle YAB = \angle AYC - \angle ABY = 90^{\circ}-\frac{C}{2} - \angle B = \frac{\angle A - \angle B}{2}\]and we are done.

Does anyone have a solution by moving points?

Let $P$ be the intersection of $BC$ and the angle bisector of $\angle{DAB}$. Then \begin{align*} \angle{PAC}&=\angle{PAD}+\angle{DAC}=\frac{1}{2}\angle{BAD}+\angle{DAC}=\frac{1}{2}\left(\angle{BAC}-\angle{DAC}\right)+\angle{DAC}\\ &=\frac{1}{2}\angle{BAC}+\frac{1}{2}\angle{DAC}=\frac{1}{2}\angle{BAC}+\frac{1}{2}\angle{CBA} \end{align*}so \[\angle{CPA}=\pi-\angle{PAC}-\angle{ACB}=\angle{BAC}+\angle{CBA}-\angle{PAC}=\angle{PAC}\]and thus $\triangle{CAP}$ is isosceles with apex $C$. Let $a=BC,b=CA,c=AB$. We apply barycentric coordinates with reference triangle $\triangle{ABC}$ so $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then since $BP=a-b$ and $PC=b$, $P=(0:b:a-b)$. In addition, $I=(a:b:c)$. Let $M_A$ be the second intersection of $AI$ with $(ABC)$ so that $M_A$ is the midpoint of arc $BC$ opposite $A$. I claim that $M_A=\left(-\frac{a^2}{b+c}:b:c\right)$. Clearly these coordinates satisfy $A,I,M_A$ collinear since $y:z=b:c$. We just need $M_A\in(ABC)$, which is true with these coordinates because \[-a^2bc-b^2c\left(-\frac{a^2}{b+c}\right)-c^2\left(-\frac{a^2}{b+c}\right)b=-a^2bc+\frac{a^2b^2c+a^2bc^2}{b+c}=0.\]Thus $M_A=\left(-\frac{a^2}{b+c}:b:c\right)$. Now, I claim that the equation of $\omega$ is \[-a^2yz-b^2zx-c^2xy+(c(a-b)y+b^2z)(x+y+z)=0.\]Plugging in $(x,y,z)=(1,0,0)$, we get that everything on the LHS is $0$ so $A$ lies on the circle. Plugging in $(x:y:z)=(a:b:c)$, we get \[-a^2bc-b^2ca-c^2ab+(c(a-b)b+b^2c)(a+b+c)=-abc(a+b+c)+abc(a+b+c)=0\]so $I$ lies on this circle. We just need $AC$ to be tangent to this circle. This is equivalent to the only solution to \begin{align*} -a^2yz-b^2zx-c^2xy+(c(a-b)y+b^2z)(x+y+z)&=0\\ y&=0\\ x+y+z&=1 \end{align*}being $(x,y,z)=(1,0,0)$. Solving this system, we require \[-b^2zx+b^2z(x+z)=b^2z(1-x)=b^2z^2=0\]so $z=0$ and $x=1$. So this circle is tangent to $AC$. Thus $\omega$ has equation as claimed. Next, I claim that $X=\left(\frac{a^2(a-b)}{b+c-a}:b^2:-c(a-b)\right)$. To do this, we verify that the coordinates lie on $(ABC)$ and $\omega$. We require \begin{align*} 0&=-a^2b^2(-c(a-b))-b^2(-c(a-b))\left(\frac{a^2(a-b)}{b+c-a}\right)-c^2\left(\frac{a^2(a-b)}{b+c-a}\right)b^2\\ &=a^2b^2c(a-b)+\frac{a^2b^2c(a-b)^2}{b+c-a}-\frac{a^2b^2c^2(a-b)}{b+c-a}\\ &=a^2b^2c(a-b)\left(1+\frac{a-b}{b+c-a}-\frac{c}{b+c-a}\right) \end{align*}which is true so these coordinates lie on $(ABC)$. And \[c(a-b)b^2+b^2(-c(a-b))=0\]so by adding to the equation of $(ABC)$, we deduce that these coordinates lie on $\omega$. Thus these coordinates are either for $A$ or $X$, but the $y$ component is nonzero so these coordinates are for $X$. So $X=\left(\frac{a^2(a-b)}{b+c-a}:b^2:-c(a-b)\right)$. Finally, I claim that $P,X,M_A$ are collinear. To do this, we compute the determinant \begin{align*} \begin{vmatrix} 0 & b & a-b \\ \frac{a^2(a-b)}{b+c-a} & b^2 & -c(a-b) \\ -\frac{a^2}{b+c} & b & c \end{vmatrix} &= \begin{vmatrix} 0 & b & a-b \\ \frac{a^2(a-b)}{b+c-a} & b^2 & -c(a-b) \\ -\frac{a^2}{b+c} & 0 & b+c-a \end{vmatrix} \\ &= \begin{vmatrix} 0 & b & a-b \\ \frac{a^2(a-b)}{b+c-a} & b(b+c) & 0 \\ -\frac{a^2}{b+c} & 0 & b+c-a \end{vmatrix} \\ &=-b \begin{vmatrix} \frac{a^2(a-b)}{b+c-a} & 0 \\ -\frac{a^2}{b+c} & b+c-a \end{vmatrix} +(a-b) \begin{vmatrix} \frac{a^2(a-b)}{b+c-a} & b(b+c) \\ -\frac{a^2}{b+c} & 0 \end{vmatrix} \\ &=-b\cdot a^2(a-b)+(a-b)\cdot a^2b\\ &=0 \end{align*}and thus $P,X,M_A$ are collinear. But $XM_A$ is the angle bisector of $\angle{CXB}$ so the angle bisectors of $\angle{DAB}$ and $\angle{CXB}$ intersect at a point on line $BC$.

Let the angle bisector of $\angle BAD$ cut $BC$ at $U$. Also, let $AD$ cut again the circumcircle of $ABC$ at $E$. Then by $\angle CAD\equiv \angle CBA$ we find that $BC$ is the angle bisector of $\angle ABE$, hence $U$ is the incenter of $ABE$. Thus $UC=CA~(=CE)$. Let $\omega$ cut $BC$ again at $Q$. Then $\angle BXQ=180^{o}-\angle AQX=180^{o}-\angle APX=\angle APC$ and $\angle XBQ=\angle ACP$, so $\Delta APC\sim \Delta XQB$. Since $CA$ is tangent to $\omega$ at $A$, we also have $\Delta APC\sim \Delta XAC$, so $\Delta XQB\sim \Delta XAC$. Then $XB/XC=BQ/AC$ or $XB/XC=BQ/CU$. Therefore, in order to prove that $XU$ is the angle bisector of $\angle BXC$, it suffices to prove that $BQ=BU$. This is however equivalent to $IU=IQ$ or $IQ=IA$, since $IU=IA$ ($CI$ is the perpendicular bisector of $AU$. But $\angle IQA=\angle IAC=\angle IAQ$ so $IQ=IA$, as required.

Since there are already a lot of solution. So I will provide just few observations like 1) CMa is parallel to IQ 2)AIPB is cyclic. 3)QDBX is cyclic. 4)CX AD meet at the circle tangent to AC passing through I. 5)IQP are collinear. Using above observations one can easily prove that CQ and MaP meet on the circumcircle of ABC. Just try to avoid circular proof. This problem is too easy as a 3rd problem. FE would have been more welcomed but anyway

EGMO 2019 P3 wrote: Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$. Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $ABC$. Prove that the angle bisectors of $\angle DAB$ and $\angle CXB$ intersect at a point on line $BC$. Notations: Let $\angle A=2a, \angle B= 2b, \angle C = 2c$ and let $Z=BI\cap \text{angle bisector of} \angle DAB, Y=CB\cap AZ$ and consider a point $W$ on extended $AX$ towards $X$. One line solution : Obviously, $\angle BZY= a, \angle DAC = 2b, \angle DAY = \angle YAB= a-b, \angle AYC = 90-c, \angle AXB = 2a$ and so, $\angle IZA = a = \angle CAI \implies AIZX$ is cyclic and thus, $\angle AIZ = 90+c =\angle ZXW$, but we also have $\angle BXW = 2c\implies \angle ZXB = 90-c = \angle AYC\implies XZYB$ is cyclic and thus, $\angle YXB = \angle YZB = a\implies \angle BXY = \angle ZXY =a$ and we are done.

The proof is based in some Claims. Claim 1: Lines $AD$ and $XC$ intersect on circle $(AXI)$. Proof: Let those two lines intersect at point $P$. Then, $\angle APX=\angle ACP+\angle PAC=\angle ACX+\angle DAC=\angle ABX+\angle B=$ $=\angle XBC=180^\circ-\angle XAC=180^\circ-\angle XAI-\angle IAC=$ $=180^\circ-\angle XAI-\angle AXI=\angle AIX,$ and so $P \in (AIX)$, as needed $\blacksquare$ Claim 2: The internal angle bisector of $\angle BAD$ and $BI$ intersect on circle $(AXI)$. Proof: Let these two lines intersect at point $S$. Then, $\angle ASI=\angle BAS+\angle ABS=\dfrac{\angle BAD}{2}+\dfrac{\angle B}{2}=\dfrac{\angle }{2}=\angle IAC=\angle AXI,$ and so $S \in (AXI)$, as desired $\blacksquare$ Claim 3: If $AS$ intersects $BC$ at $T$, then quadrilateral $BXST$ is cyclic. Proof: Indeed, note that $\angle ATC=\dfrac{\angle BAD}{2}+\angle B=90^\circ-\dfrac{\angle C}{2}$ and $\angle BXS=\angle BXA-\angle AXS=(180^\circ-\angle C)-(180^\circ-\angle AIB)=90^\circ-\dfrac{\angle C}{2},$ and so the two angles are equal, i.e. $BXST$ is cyclic $\blacksquare$ Now, to finish, we have $\angle BXT=\angle BST=\angle ASI=\angle AXI=\angle IAC=\dfrac{\angle A}{2}=\dfrac{\angle BXC}{2},$ and so $XT$ bisects $\angle BXC$, hence we may finish.

Let $CX\cap \omega$ at $Y$.$\angle YXA=\angle CAY=\angle CAD$ so $A,Y,D$ collinear.And $CA^2=CY\cdot CX=CD\cdot CB$ which implies $XYDB$ is cyclic.Denote by $M$ the midpoint of arc $CB$,$T=MX\cap CB$.Do some angle chasing and notice that to finish the problem $AITB$ should be cyclic which is easy to prove.

Let $\angle BAC=2a$, $\angle CBA=2b$, and $\angle ACB=2c$. Let $Y$ be the intersection of the angle bisector of $\angle CXB$ with $BC$. Note that; \[\angle YXB=\frac{1}{2}\angle CXB=\frac{1}{2}\angle CAB=\angle CAI=\angle AXI,\]by cyclic and tangent properties. Also, we have that \[\angle YBX=\angle ABC+\angle ABX=2b+(180-(\angle BAX+\angle AXB))=2b+(180-(x+(180-\angle ACB)))\]\[=2b+(180-(x+180-2c))=2b+2c-x=180-(2a+x)=180-(a+(a+x))\]\[=180-(\angle AXI+\angle IAB+\angle BAX)=180-(\angle AXI+\angle IAX)=\angle AIX,\]which, combined with the first angle chase, gives that $\triangle AIX$ and $\triangle YBX$ are spirally similar around $X$. This means that if we let $M=AY\cap IB$, we have that $AIMX$ and $MXBY$ are cyclic. Using this, we have that \[\angle MAB=180-(\angle AMB+\angle MBA)=\angle YMB-\angle IBA=\angle YXB-\frac{1}{2}CBA=a-b,\]which is $\frac{1}{2}$ of $\angle CAB-\angle ABC$, implying that $AY$ is indeed the angle bisector of $\angle DAB$, finishing the problem.

Simple geometry. Let $CX\cap AD=T$. $\angle CBA=\angle CXA=\angle CAT$ gives that $T$ lies on $\omega$. Since $\angle ADC=\angle CAB=\angle CXB$ we have $DTXB$ is cyclic. Let angle bisector of $\angle CXB$ meets $BC$ at point $N$. $\angle IXA=\angle IAC=\angle BAC/2=\angle CXB/2=\angle NXB$ and $\angle NBX=\angle CTD=\angle ATX=\angle AIX$, thus $\Delta XIA\sim \Delta XBN$. Then $\Delta XAN\sim \Delta XIB$, rest is angle chasing.

Let the bisector of $\angle CXB$ meet $BC$ at $X'$ Let the bisector of $\angle BAD$ meet $BC$ at $A'$ Let the $AI$ meet $BC$ at $I'$ Claim: $AXX'I'$ is cyclic Observe $\measuredangle AXX'=\measuredangle AI'X'$- Claim:$ BXII' $is cyclic Observe $\measuredangle IXB=\angle AXB - \angle AXI $ We know that $\angle AXI = \angle CAI$ Therefore we get $\measuredangle IXB=\measuredangle BI'I$- Claim: $AXX'I'$ is cyclic Observe $\measuredangle AXI'=\measuredangle AA'I'$- Therefore we get $ X'=A' $ We are done Easy for a P3

By $\sqrt{bc}$ inversion the problem becomes: Quote: In triangle $ABC$ with $A$-excenter $I_A$, let $M=\overline{AI_A} \cap \overline{BC}$, let $X$ be the point on $\overline{BC}$ with $\overline{XI_A} \parallel \overline{AB}$. Prove that if $(ALX)$ intersects $(ABC)$ again at $T$, then $T$ is the midpoint of arc $ACB$. Let $L$ be the midpoint of arc $BC$. Then $\measuredangle AMC=\measuredangle ABL$ so it suffices to show that $T,X,L$ are collinear. By extraversion about $A$ and then relabeling $B$ to $A$, the problem becomes Quote: In triangle $ABC$ let $D$ and $E$ be the midpoints of arcs $AB$ and $AC$. Let $\overline{DE}$ intersect $\overline{AB}$ at $P$. Prove that $\overline{IP} \parallel \overline{AC}$. Let $\overline{DE}$ intersect $\overline{AC}$ at $Q$. It is well-known that $A$ and $I$ are reflections of each other about $\overline{DE}$, so since $\overline{AB}$ and $\overline{AC}$ are symmetric about $\overline{AI}$ we have $AP=AQ=IP=IQ$, hence $APIQ$ is a rhombus and the desired parallelism follows. $\blacksquare$

Somehow nobody has complex bashed this (despite not one but two bary sols above) so I'll do that. Let $(ABC)$ be the unit circle with $A=x^2,B=y^2,C=z^2$ such that the midpoints of arcs $BC, CA, AB$ are $-yz,-zx,-xz$ respectively, so $I=-xy-yz-zx$. Let $\overline{AD}$ intersect $(ABC)$ again at $P$. The angle condition implies that $\measuredangle APC=\measuredangle ABC=\measuredangle CAP$, so $C$ is the midpoint of arc $AP$ and thus $P=\tfrac{z^4}{x^2}$. Let $M$ be the midpoint of arc $BP$ not containing $A$. Then $M=\tfrac{yz^2}{x}$, since its antipode forms an isosceles trapzoid with $A$, $C$, and the midpoint of arc $AB$. Thus we can calculate $Q:=\overline{BC} \cap \overline{AM}$ as $$\frac{xyz^2(y^2+z^2)-y^2z^2(x^2+\frac{yz^2}{x})}{xyz^2-y^2z^2}=\frac{x^2(y^2+z^2)-y(x^3+yz^2)}{x(x-y)}=\frac{xz^2+yz^2-x^2y}{x}=z^2+\frac{yz^2}{x}-xy.$$ According to the tangency fact, we should have $\measuredangle CAI=\measuredangle AXI$, so $\overline{XI}$ intersects $(ABC)$ again at $-\tfrac{x^2y}{z}$. Thus we may calculate $X$ as $$\frac{-\frac{x^2y}{z}-(-xy-yz-zx)}{-\frac{x^2y}{z}(-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx})-1}=\frac{z^2(xy+yz+zx)-x^2yz}{xz+x^2+xy-z^2}.$$On the other hand, the line joining $Q$ with the midpoint of arc $BC$ intersects $(ABC)$ again at $$\frac{-yz-(z^2+\frac{yz^2}{x}-xy)}{-yz(\frac{1}{z^2}+\frac{x}{yz^2}-\frac{1}{xy})-1}=\frac{z^3x+yz^3+xyz^2-x^2yz}{xy+x^2-z^2+xz},$$which is just $X$, i.e. $\overline{XQ}$ bisects $\angle CXB$ as desired. $\blacksquare$

Let $P$ be the intersection of the bisector of $\angle BAD$ with $BC$, and $Q$ be the same for $\angle BXC$. We will show that $$\frac{BP}{CP}=\frac{BQ}{CQ},$$which would imply $P=Q$. \begin{claim} We have $$\frac{BP}{CP}=\frac{c}{b}\cdot\frac{\sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}.$$\end{claim} This follows by Law of Sines/Ratio lemma on $\triangle ABC$ and $P$, since $$\angle ABP=\frac{1}{2}\angle ABD=\frac{1}{2}(\alpha-\beta).$$ Now, we can essentially get rid of $D$ from the problem, and just focus on $X$. Note that by Angle Bisector Theorem, $$\frac{BQ}{CQ}=\frac{BX}{CX}.$$ \begin{claim} We have $$\frac{XB}{XC}=\frac{c}{b}\cdot\frac{a-b}{c}.$$\end{claim} Since there are two circles passing through both $A$ and $X$, this motivates inverting around $A$. Perform a $\sqrt{bc}$ inversion followed by a reflection across the angle bisector. We have $$\frac{XB}{XC}=\frac{r^2\cdot \frac{X'B'}{AX'\cdot AB'}}{r^2\cdot \frac{XC'}{AX'\cdot AC'}}$$$$=\frac{X'B'}{X'C'}\cdot \frac{AC'}{AB'}$$$$=\frac{X'C}{X'B}\cdot \frac{AB}{AC}=\frac{X'C}{X'B}\cdot \frac{c}{b}.$$ Note that the inverse of $(AIX)$ is a line through $I_a$ parallel to $AB$. Thus, $X'$ is the point on $BC$ such that $I_AX'\parallel AB.$ Use barycentric coodinates. If $I_a=(-a:b:c)$, then going along parallel to $AB$ we lose $y$ at the same rate we gain $x$, so $X'=(0:-a+b:c)$. This directly implies that $$\frac{X'C}{X'B}=\frac{a-b}{c},$$this showing the claim. Only one thing left to do... \begin{claim} We have $$\frac{a-b}{c}=\frac{\sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}.$$\end{claim} Of course, convert this to $$\frac{\sin\alpha-\sin\beta}{\sin\gamma}=\frac{\sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}.$$ Let $$\theta_1=\frac{\alpha+\beta}{2},\theta_2=\frac{\alpha-\beta}{2},$$so the equation becomes $$\frac{\sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)}{\sin180-2\theta_1}=\frac{\sin\theta_2}{\sin\theta_1}$$ However, $$\frac{\sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)}{\sin180-2\theta_1}=\frac{2\cos\theta_1\sin\theta_2}{\sin2\theta_1}$$ $$=\frac{2\cos\theta_1\sin\theta_2}{2\sin\theta_1\cos\theta_1}=\frac{\sin\theta_2}{\sin\theta_1},$$as desired.

Cyclic Quad Question Let $F = CX \cap (IXA)$ , $H$ be feet of angle bisector of $\angle BAD$, $G = H \cap IB$ Claim 1 : $\overline{A-F-D}$ $$\angle AFC = \angle AFI + \angle IFC = \frac{\angle A}{2} + \angle XAI = \angle XAC = 180 - \angle XBC$$from power of point $$CD.CB = CA^2 = CF.CX$$ we get $X,B,D,F$ are cyclic and hence we get $\overline{A-F-D}$ Claim 2 : $X,G,H,B$ cyclic from $\angle BGH = \frac{\angle B}{2} + \frac{\angle A - \angle B}{2} = \angle AXI$ which give us $X,G,I,A$ are cyclic $$\angle XGA = \angle XFA = 180 - \angle AFC = \angle XBC$$ and hence $\frac{\angle A}{2} = \angle AGI = \angle BGH = \angle BXH$ and from $\angle BXC = \angle BAC$ we get $XH$ is angle bisector of $\angle XHC$ $\blacksquare$

Assume WLOG $2\angle B < \angle A$ for config purposes. Suppose $XI \cap (ABC) = P$. We can angle chase \[\angle DAI = \angle CXI = \angle CBP = \frac A2 - B, \quad \angle PBI = \frac{A-B}{2}, \angle PCI = 90 + \frac B2.\] Suppose the angle bisectors of $\angle DAB$, $\angle CXB$ intersect $BC$ at $X_1$, $X_2$. Ratio lemma gives us \[\frac{BX_1}{X_2C} = \frac{AB \sin \frac{A-B}{2}}{AC \sin \frac{A+B}{2}} = \frac{\sin C \sin \frac{A-B}{2}}{\sin B \cos \frac C2} = \frac{\sin \frac C2 \sin \frac{A-B}{2}}{\sin \frac B2 \cos \frac C2}.\] Angle bisector theorem and LOS then tells us \[\frac{BX_2}{X_2C} = \frac{BX}{XC} = \frac{\sin \angle BPI}{\sin \angle CPI} = \frac{BI}{CI} \cdot \frac{\sin \frac{A-B}{2}}{\sin 90+\frac B2} = \frac{\sin \frac C2 \sin \frac{A-B}{2}}{\sin \frac B2 \cos \frac B2}. \quad \blacksquare\]

Indeed easy for a P3 and significantly easier than the combi P2 (although I guess it being P2 hints that the solution there should be quick and clean with a known technique). I see great solutions which use angles only (but have quite a bit of additional points), here is the one which is most natural to me - intersect bisectors at the common midpoint of arc and then chase angles and similarities. Firstly, if the angle bisector of $\angle DAB$ intersects $BC$ at $K$, then $\angle CAK = \angle CAD + \angle DAK = \angle CAD + \frac{1}{2}\angle DAB = \angle CAD + \frac{1}{2}\angle CAB - \frac{1}{2}\angle CAD = \frac{1}{2}(\angle CAB + \angle ABC) = 90^{\circ} - \frac{1}{2}\angle ACB$ and thus $CA = CK$. So the real problem here is as follows: Let $ABC$ $(BC > AC)$ be a triangle with circumcircle $\omega$ and incentre $I$. A circle through $A$ and $I$ tangent to $AC$ intersects $\omega$ for the second time at $X$. The angle bisector of $\angle CXB$ intersects the line $BC$ at $L$. Prove that $AC = CL$. To solve this, let $AI$ and $XL$ intersect on the midpoint $M$ of the arc $\widehat{BC}$ of $\omega$. We have $\angle MBL = \frac{1}{2}\angle BAC = \frac{1}{2}\angle BXC = \angle BXM$ and so $\triangle MBL \sim \triangle MXB$, giving $MB^2 = ML \cdot MX$. On the other hand, $MB = MI$ by the Trillium lemma, hence $MI^2 = ML \cdot MX$. Together with $\angle IML = \angle XMI$ we obtain $\triangle IML \sim \triangle IXM$ and thus $\angle MIL = \angle IXL$. Now to employ the tangency, we have $\angle IXL = \angle AXB - \angle LXB - \angle AXI = 180^{\circ} - \angle ACB - \frac{1}{2}\angle BAC - \angle CAI = \angle ABC$. Hence $\angle CIL = \angle CIM + \angle MIL = \frac{1}{2}(\angle BAC + \angle ACB) + \angle IXL = 90^{\circ} + \frac{1}{2}\angle ABC = \angle AIC$, which together with $\angle ACI = \angle LCI$ and the common side $CI$ implies $\triangle ACI \cong \triangle LCI$ and $AC = CL$, as desired.

Let the angle bisector of $\angle DAB$ intersect BC at $X_1$ and let the angle bisector of $\angle CXB$ intersect BC at $X_2$ $\Rightarrow$ we want to show that $X_1 \equiv X_2$ $\Rightarrow$ we will prove that $\frac{BX_1}{X_1C} = \frac{BX_2}{X_2C}$. Let $XI \cap (ABC) = P$. Now by some angle chase we get that $\angle DAI = \angle CXI = \angle CBP = \frac{\alpha}{2} - \beta$, also $\angle PBI = \frac{\alpha-\beta}{2}$, also $\angle DAX_1 = \frac{\alpha-\beta}{2}$ and $\angle PCI = 90 + \frac{\beta}{2}$. Now by the ratio lemma we get that $\frac{BX_1}{X_1C} = \frac{AB}{AC}.\frac{\sin\angle BAX_1}{\sin\angle X_1AC} = \frac{2R\sin\gamma}{2R\sin\beta}.\frac{\sin(\frac{\alpha-\beta}{2})}{\sin(\frac{\alpha+\beta}{2})} = \frac{\sin\gamma}{\sin\beta}.\frac{\sin(\frac{\alpha-\beta}{2})}{\sin(\frac{\alpha+\beta}{2})} = \frac{2\sin\frac{\gamma}{2}.\cos\frac{\gamma}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}. \frac{\sin(\frac{\alpha-\beta}{2})}{\sin(\frac{\alpha+\beta}{2})} = \frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta}{2}}.\frac{\sin(\frac{\alpha-\beta}{2})}{\cos\frac{\beta}{2}}$. Now $\frac{BX_2}{X_2C} = \frac{BX}{CX} = \frac{2R.\sin \angle XCB}{2R.\sin \angle CBX} = \frac{\sin \angle IPB}{\sin \angle CPI} = \frac{BI}{CI}.\frac{\sin \angle PBI}{\sin \angle PCI} = \frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta}{2}}.\frac{\sin(\frac{\alpha-\beta}{2})}{\sin(90 + \frac{\beta}{2})}$ $\Rightarrow$ $\frac{BX_1}{X_1C} = \frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta}{2}}.\frac{\sin(\frac{\alpha-\beta}{2})}{\cos\frac{\beta}{2}}$ and $\frac{BX_2}{X_2C} = \frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta}{2}}.\frac{\sin(\frac{\alpha-\beta}{2})}{\sin(90 + \frac{\beta}{2})}$ $\Rightarrow$ we only need to show that $\cos\frac{\beta}{2} = \sin(90 + \frac{\beta}{2})$, which is obviously true $\Rightarrow$ $\frac{BX_1}{X_1C} = \frac{BX_2}{X_2C}$ and $X_1 \equiv X_2$ $\Rightarrow$ we are ready.

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I could spend more time trying to find a nice solution but like why bother? Let $P = \overline{CX} \cap \Gamma$ and notice that $XPDB$ is cyclic by radical axis, so $P$ lies on $\overline{AD}$ as $\angle ADC = \angle BXC = \angle PDC$. Now we bash in the most brainless way possible. Compute the $r_\Gamma = \frac{AI}{\sin \frac A2}$ and $r_{(ABD)} = \frac{AD}{\sin B}$. As a homothety at $A$ with ratio $\frac{r_{(ABD)}}{r_\Gamma}$ takes $P$ to $D$, we can compute \[PD = AD\left(1-\frac{r_{\Gamma}}{r_{(ABD)}}\right) = AD \left(1-\frac{AI \sin B}{AD \sin \frac A2}\right).\]Let $K$ be the foot of the angle bisector of $\angle BAD$. Now using similarity to compute $AD$ and law of sines to compute $AI$, we have \begin{align*} \frac{XB}{XC} = \frac{PD}{CD} &= \frac{\frac{\sin B \sin C}{\sin A} - \frac{2 \sin B \sin \frac B2 \sin \frac C2}{\sin \frac A2}}{\frac{\sin^2 B}{\sin A}} \\ &= \frac{\sin C - 4 \cos \frac A2 \sin \frac B2 \sin \frac C2}{\sin B} \\ &= \frac{2\sin \frac C2\left(\cos \frac C2 - 2 \cos \frac A2 \sin \frac B2\right)}{\sin B} \\ &= \frac{2\sin \frac C2 \sin\left(\frac{A-B}2\right)}{\sin B} \\ &= \frac{\sin C \sin \left(\frac{A-B}2\right)}{\sin B \sin \left(\frac{A+B}2\right)} = \frac{BK}{CK}. \end{align*}Thus done.

This solution uses the fact that the midpoint of the arc $BC$ not containing $A$ is the center of $(BIC)$. Let $J$ be the point on segment $BC$ such that $CJ=CA$ and $E$ be the second intersection of $AD$ with $(ABC)$. Label the midpoints of arcs $AB$ and $BC$ not containing $C$ and $A$ with $M_C$ and $M_A$, respectively, and let $X'\neq M_C$ be the second intersection of $(ABC)$ and $(M_CJI)$. First, notice that $\angle JAB=\angle BAC-\angle JAC=\angle BAC-\left(90^\circ-\frac12\angle ACB\right)=\frac12\angle BAD$, implying $J$ is the point where the angle bisector of $\angle DAB$ intersects $BC$. Also, since $BC$ bisects $\angle ABE$ and $CA=CJ$, $J$ is the incenter of $\triangle ABE$. Claim: $X'$ lies on $JM_A$. Proof: Since $J$ is the incenter of $\triangle ABE$, $M_CJ=M_CA=M_CI$. We now have \begin{align*} \angle M_CX'J&=\angle M_CIJ\\ &=90^\circ-\frac12\angle IM_CJ\\ &=90^\circ-\frac12\angle ABC\\ &=\frac12\angle BAC+\frac12\angle ACB\\ &=\angle M_CX'M_A. \end{align*}This gives $\angle M_AX'I=\angle JX'I=\angle JM_CI=\angle ABC$, so $\angle IXA=\frac12\angle BAC$, which implies that $X'=X$, as desired.