Let $\ell$ denote the angle bisector of $C$. Let $A_1$ and $B_1$ be the reflections of $A$ and $B$ in $\ell$, respectively. Set $Q_1 = CQ \cap AB$. We wish to show $CP$ and $CQ_1$ are isogonal. Note that by homothety, $Q_1$ is the tangency point of the $C$-excircle on side $AB$ of $\triangle ABC$. Consider an inversion which sends $A$ to $B_1$ and $B$ to $A_1$. Then $\omega$ is sent to the $C$-excircle of $\triangle CA_1B_1$ and $\Omega$ is sent to $A_1B_1$. So $P^\ast$ is the tangency point of the $C$-excircle of $\triangle A_1B_1C$ to side $A_1B_1$. In other words, $P^\ast$ is the reflection of $Q_1$ over $\ell$. Therefore, $CP$ is isogonal to $CQ_1$ as desired.

Let $AB$ and $AC$ touch the circle $\omega$ at $X$ and $Y$ respectively and let $W$ be the midpoint of $XY$ .It is well known that $PX$ bisects $\angle BPA$. .Also it is well known that $PA$ is the $P$-symmedian of $\Delta PXY$ .Hence $\Rightarrow \angle WPY=\angle APX=\angle XPB \Rightarrow \angle BPW=\angle YPX =\angle AXY$ .Similarly $\angle WPC=\angle WYA \Rightarrow PW$ bisects $\angle BPC$ Let $BP$ and $PC$ intersect $\omega$ at $K$ and $L$ respectively Hence Line $PW$ bisects the arc $KL$ not containig $P$ But since $BC||KL$ it also bisects arc $DE$ not containing $P$ where $D$ and $E$ are the intersections of $BC$ with $\omega$.Hence if $PW$ intersects $\omega$ again at $J$ we have $J\equiv Q$ .Now Let $AP$ intersect $\omega$ again at $Q'$.We have $\angle AXQ'=\angle XPA=\angle QPY=\angle QYA$ .Hence it is easy to get the desired result edit-A small mistake .I have solved the problem considering the A-mixtilinear incircle but not that of C

Let $\omega \cap CA = E$ and $\omega \cap CB = F$. It is well known that the incenter $I$ is the midpoint of $EF$. Let $PE, PF, PI$ intersect $\Omega$ at $E', F', I'$. Now $\angle APE = \angle FPI = \angle F'PI'$ ($PA$ symmedian). It is also well known that $E', F'$ are the arc midpoint of $CA, CB$ respectively in $\Omega$. Now $\frac 12 \angle B= \angle APE' = \angle F'PI'$ implies $I'$ is the arc midpoint of $AB$ (not containing $P$). Let $PI'$ intersect $\omega$ at $I''$. Then, if $O' = \text{ center of } \omega$ and $O = \text{ center of } \Omega$, then $I'O || I''O'$, so $I''O' \perp BC$, or tangent at $I''$ to $\omega$ is parallel to $BC$. So, $I'' = Q$ Now, midpoint of $EF$ or $I$ is the inversion of $C$ wrt $\omega$, and $I$ is on $PQ$, which implies $CPO'Q$ cyclic. Again, $O'P = O'Q$ implies $CO'$ bisects $\angle PCQ$. As it is well known that $CO'$ is the bisector of $\angle BCA$, $CP$ and $CQ$ is symmetric wrt $CO'$, so $\angle ACP =\angle BCQ$[proved].

Let $D$ and $E$ be the midpoints of arcs $AB$ and $ACB$ on $\Omega$, respectively. Let $I$ be the incenter of triangle $ABC$, so $I$ lies on $CD$. From the post http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=5131, $I$ lies on $PE$. There exists a homothety $h$ centered at $P$ taking $\Omega$ to $\omega$. From the definition of $Q$, $h(E) = Q$. [asy][asy] pair excentre(pair A, pair B, pair C) { return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B))); }; pair mixtilinearcirclecenter(pair A, pair B, pair C) { pair I, D, E; I = incenter(A,B,C); D = extension(A, B, I, rotate(90)*(A - I) + I); E = extension(A, C, I, rotate(90)*(A - I) + I); return(extension(D, rotate(90)*(A - D) + D, E, rotate(90)*(A - E) + E)); }; real mixtilinearcircleradius(pair A, pair B, pair C) { pair I, D, E, O; I = incenter(A,B,C); D = extension(A, B, I, rotate(90)*(A - I) + I); E = extension(A, C, I, rotate(90)*(A - I) + I); return(abs(mixtilinearcirclecenter(A,B,C) - D)); }; path mixtilinearcircle(pair A, pair B, pair C) { return(Circle(mixtilinearcirclecenter(A,B,C),mixtilinearcircleradius(A,B,C))); }; pair mixtilinearcircletangentpoint(pair A, pair B, pair C) { pair[] R; R[0] = abs(circumcenter(A,B,C) - A); R[1] = mixtilinearcircleradius(A,B,C); return((R[0]*mixtilinearcirclecenter(A,B,C) - R[1]*circumcenter(A,B,C))/(R[0] - R[1])); }; unitsize(0.4 cm); pair A, B, C, Cp, D, E, F, G, O, P, Q, R; pair[] I; A = (0,0); B = (14,0); C = (2,12); I[0] = incenter(A,B,C); I[1] = excentre(A,B,C); I[2] = excentre(B,C,A); I[3] = excentre(C,A,B); D = (I[0] + I[3])/2; E = (I[1] + I[2])/2; P = mixtilinearcircletangentpoint(C,A,B); Q = intersectionpoint(I[0]--E,mixtilinearcircle(C,A,B)); O = circumcenter(A,B,C); Cp = intersectionpoint(interp(P,C,0.1)--C,mixtilinearcircle(C,A,B)); draw(incircle(A,B,C),gray(0.7)); draw(mixtilinearcircle(C,A,B),red); draw(E--C--D); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(C--P--E); draw(C--Q--Cp); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NW); dot("$C'$", Cp, SE); dot("$D$", D, S); dot("$E$", E, N); dot("$I$", I[0], dir(0)); dot("$P$", P, SW); dot("$Q$", Q, SE); dot(mixtilinearcirclecenter(C,A,B)); [/asy][/asy] Let $C' = h(C)$. Then $C'Q$ is parallel to $CE$. Since $DE$ is a diameter of $\Omega$, $CE$ is perpendicular to $CD$. Hence, $C'Q$ is perpendicular to $CD$ (which passes through the center of $\omega$). It follows that $\angle ACP = \angle QCB$.

very easy: let $CA$ touch $\omega$ at $X$ consider the composition of an inversion with center $C$ and radius $CX$ and the symmetry wrt the bisector of $\angle ACB$. it maps circle $ACB$ into $l$ and circle $\omega$ into itself so it maps $P$ to $Q$ therefore $\angle ACP=\angle BCQ$.

leader wrote: very easy: let $CA$ touch $\omega$ at $X$ consider the composition of an inversion with center $C$ and radius $CX$ and the symmetry wrt the bisector of $\angle ACB$. it maps circle $ACB$ into $l$ and circle $\omega$ into itself so it maps $P$ to $Q$ therefore $\angle ACP=\angle BCQ$. Why the circumcircle of ABC becomes $l$ (i guess it is the angle bisector of the angle in C)?

bozzio wrote: Why the circumcircle of ABC becomes $l$ (i guess it is the angle bisector of the angle in C)? No, $\ell$ is as defined in the problem. Writing out the details ... it's easy to see $\omega$ is fixed. $\Omega$ gets mapped by the inversion to a line tangent to $\omega$ and antiparallel to $AB$ through $\angle C$. Then reflection takes it through a line also tangent to $\omega$, but now parallel to $AB$. This line is $\ell$. leader wrote: very easy Only if you know inversion

Let $AC \cap \omega=X, BC \cap \omega=Y, PX \cap \Omega=M, PY \cap \Omega=N(M,N \neq P)$. It is well-known that $M,N$ are midpoints of $\widehat {AC},\widehat {AB}$ in $\Omega$ and $I$, the incircle of $\triangle ABC$ is the midpoint of $XY$. $\angle AIX=\angle AIC-\angle CIX=90^{\circ}+\frac 12\angle ABC-90^{\circ}=\angle APX \Rightarrow AXIP$ is concyclic. Similarly $BYIP$ is concyclic. Now $\angle CXI=\angle CIY \Rightarrow \angle API=\angle BPI$. Let $PA \cap \omega=D,PB \cap \omega=E,AB \cap \omega=R,S$.($R$ is closer to $A$ than $S$).$PI \cap \omega=Q' \neq P$. So $\angle DPQ'=\angle EPQ' \Rightarrow \widehat{DQ'}=\widehat{EQ'}$. From homothetic triangles $PAB,PDE$ we get $DE \parallel RS \Rightarrow RDES$ is an isosceles trapezoid. Hence $\widehat{RD}=\widehat {SE}$. Hence $\widehat {RQ'}=\widehat {SQ'}$ which implies the tangent at $Q'$ is parallel to $AB$. So $Q=Q'$. Let $O$ be the circumcentre of $\omega$. $CXOY$ is cyclic. Now $IP.IQ=IX.IY=IO.IC \Rightarrow QIPC$ is cyclic. Since $OP=OQ$ we conclude $CI$ bisects $\angle PCQ$ meaning $CP,CQ$ are isogonals of $\angle ACB$ from where the result follows.

v_Enhance wrote: No, $\ell$ is as defined in the problem. Writing out the details ... it's easy to see $\omega$ is fixed. $\Omega$ gets mapped by the inversion to a line tangent to $\omega$ and antiparallel to $AB$ through $\angle C$. Then reflection takes it through a line also tangent to $\omega$, but now parallel to $AB$. This line is $\ell$. Ok I love this solution. A couple of things which I still don't understand. First inversion trasform circle which don't pass to O in circle; so $\omega$ is trasformed in an other circle, X and Y (defined as X but on BC) are fixed, now don't we need a third point to say that $\omega$ is fixed? Which point is it? Second, $\Omega$ becomes a line thorough the points of intersections between $\Omega$ and the circle wrt we are appling inversion, how do we say it is tangent to $\omega$ and is antiparallel to AB? I know these are stupid questions but I really like this solution and I'd like to understand it Thamk you for your help

A sketch of my solution. Let $ \omega $ touch $AC$ and $ BC$ at $ X $and $ Y $respectively. First prove that $ PQ$ intersects $XY $in its middle $ I$, which is incenter of triangle $ ABC$. Then $CP$ is a symmedian of $ XYP$ so $\angle $$ XPC$ is equal to $\angle $ $IPY$. Then using this fact and sine law we conclude that triangles $PXC$ and $ YQC$ are similar from which the result follows.

bozzio wrote: First inversion trasform circle which don't pass to O in circle; so $\omega$ is trasformed in an other circle, X and Y (defined as X but on BC) are fixed, now don't we need a third point to say that $\omega$ is fixed? Which point is it? Second, $\Omega$ becomes a line thorough the points of intersections between $\Omega$ and the circle wrt we are appling inversion, how do we say it is tangent to $\omega$ and is antiparallel to AB? I know these are stupid questions but I really like this solution and I'd like to understand it Nah, glad to help. I will let denote the inverse of a point, line, etc. with a star. $\omega^\ast = \omega$ because the circle of inversion is orthogonal to it. In other words, consider two points $P$ and $Q$ on $\omega$ such that $PQ$ passes through $C$. Then $CP \cdot CQ = CX^2$ by Power of a Point, and yet $CX$ is the radius of inversion, hence $P$ and $Q$ map to one another. Therefore, every point on $\omega$ lies on $\omega^\ast$ and vice-versa. $\Omega^\ast$ is tangent to $\omega^\ast$ because $\Omega$ and $\omega$ are tangent. Inversion preserves incidence. Since $\Omega$ and $\omega$ intersect at exactly one point, $\Omega^\ast$ and $\omega^\ast$ do as well The fact that $\Omega^\ast$ is antiparallel to $AB$ follows from the fact that $A,B,A^\ast,B^\ast$ are concyclic and $\Omega^\ast$ is the line through $A^\ast$, $B^\ast$. bozzio wrote: Ok I love this solution. Hmm I was actually rather disappointed when I found it. (leader's solution is basically the same as mine above, rewritten so as to avoid the extra point $Q_1$ that I used.) The point $Q$ feels very artificial, and the very first thing I did was get rid of it by considering $Q_1$ instead. Once you consider the excircle, the inversion becomes the "obvious" thing to do because it allows you to basically swap the circles, and after inversion the conclusion is obvious. In fact, if I were to guess, I would imagine the problem author started with the inversion, made it into a problem, and then added the point $Q$ just to make it cleaner/harder. So for a contest problem I don't like it so much -- it's too easy with inversion, and too difficult without (indeed, both angles are quite hard to deal with).

Here I will post my solution First, note that $P$ is the center of homothety that brings $\omega$ to $\Omega$. Let the line $PC$ intersects $\omega$ at a point $C'$. Draw the tangent line from $C'$ in $\omega$, name it $l$, and the tangent line from $C$ in $\Omega$. Surely both of them are parallel to each other. Then we reflect the whole picture by the bisector of $\angle ACB$. Since $\omega$ is tangent to both $CA$ and $CB$, then $\omega$ will be brought to $\omega$ again (or in the other word, $\omega$ will remain constant). The tangent line from $C$ in $\Omega$ will be brought into the line parallel to $AB$ passing through $C$. That implies the line $l$ reflected will be parallel to $AB$. As the reflection preserve the number of intersection point between 2 objects, the line $l$ will still be tangent to $\omega$ Note that the line tangent to $\omega$ parallel to $AB$ meets $\omega$ at $Q$. Hence we know that $CP$ will be brought to $CQ$ by that reflection. That means $CP$ and $CQ$ are isogonal to each other, hence completes the proof.

This problem is equivalent to the fact that the Nagel point is isogonal conjugates with $X_{56}$, the exsimilicenter of (O) and (I). The latter is the intersection point of $CP$ with its analogous lines for the other vertices, by Monge's theorem on (O), (I), and each of the mixtilinear incircles. The former is on $CQ$ because a homothety about $C$ takes $Q$ and the $C$-mixtilinear incircle to the extangent point on $AB$ and the $C$-excircle. Proving the conjugacy is done easily with the inversion v_Enhance did. Of course, this was all that was needed for the problem.

Let's call $D, E, \omega$ the midpoints of the arcs $\stackrel{\frown}{AB}, \stackrel{\frown}{ACB}$ and the center of $\omega$. As already pointed out by nsato, $P-I-Q-E$ are collinear, and so are $C-I-\omega - D$. Clearly $CEDP$ is cyclic, and so is $CQ\omega P$, since $\omega P\parallel DE$; with $\omega P=\omega Q$ we are done, since $C\omega$ is the angle bisector of $\angle PCQ$. Best regards, sunken rock

Lemmas: 1. If circle $\gamma$ is internally tangent to circle $\omega$ at point $P$, and chord $AB$ on $\omega$ is tangent to $\gamma$ at $M$, $PM$ bisects $\overarc{AB}$. Simply take the intersections of $PA$ and $PB$ with $\gamma$. We know that these two new points will form a line parallel to the original, and it is easy to see that these points with $M$ form an isosceles triangle. Taking the homothety from $\gamma$ to $\omega$ yields the desired result. 2. $I$ is the midpoint of the touch-chord of $\gamma$ on $CA, CB$. Use the previous lemma and Pascal's directly. 3. $PI$ bisects $\overarc{AB}$ Consider that by the first lemma $PX$ and $PY$ are bisectors of $\angle APC$ and $\angle BPC$ respectively. Then we also know $PA$ is the symmedian of $\triangle PXY$ because $AX$ and $AY$ are tangents. Then \[\angle IPB=\angle IPY+\angle APY=\angle IPY+\angle XPC=\angle XPY\] and similarly $\angle IPC=\angle XPY$, thus $\angle IPB=\angle IPC$ and this implies the arc equality. Now consider the points $X,Y$ on $CA$ and $CB$ respectively such that $\omega$ is tangent to $AB$ and $BC$. Then we know $\angle XPA=\frac{\angle C}{2}$ by the previous lemma. Let the tangent of $Q$ to $\omega$ intersect $CB$ at $Z$. Then we know since it's parallel to $AB$ $\angle QZY= 180-\angle C$, and since $QZ=ZY$, $\angle ZYQ=\frac{\angle C}{2}=\angle XPA$. By the first lemma we know $PQ$ will bisect $\overarc{AB}$ (since parallel), thus $PIQ$ are collinear. Them we see by extending line $CQ$ that it is the symmedian of $\triangle YQX$, thus $180-\angle CQY=\angle IQX=\angle PXA=180-\angle CSP$, and $\angle CQY=\angle PXC$, implying with the last angles $\angle PCA=\angle QCB$.

Took me forever to notice, but... nsato wrote: Let $D$ and $E$ be the midpoints of arcs $AB$ and $ACB$ on $\Omega$, respectively. Let $I$ be the incenter of triangle $ABC$, so $I$ lies on $CD$. From the post http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=5131, $I$ lies on $PE$. There exists a homothety $h$ centered at $P$ taking $\Omega$ to $\omega$. From the definition of $Q$, $h(E) = Q$. Let $C' = h(C)$. Then $C'Q$ is parallel to $CE$. Since $DE$ is a diameter of $\Omega$, $CE$ is perpendicular to $CD$. Hence, $C'Q$ is perpendicular to $CD$ (which passes through the center of $\omega$). It follows that $\angle ACP = \angle QCB$. I don't think we need $I$ at all in this solution: it stands perfectly well without it. [asy][asy] pair excentre(pair A, pair B, pair C) { return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B))); }; pair mixtilinearcirclecenter(pair A, pair B, pair C) { pair I, D, E; I = incenter(A,B,C); D = extension(A, B, I, rotate(90)*(A - I) + I); E = extension(A, C, I, rotate(90)*(A - I) + I); return(extension(D, rotate(90)*(A - D) + D, E, rotate(90)*(A - E) + E)); }; real mixtilinearcircleradius(pair A, pair B, pair C) { pair I, D, E, O; I = incenter(A,B,C); D = extension(A, B, I, rotate(90)*(A - I) + I); E = extension(A, C, I, rotate(90)*(A - I) + I); return(abs(mixtilinearcirclecenter(A,B,C) - D)); }; path mixtilinearcircle(pair A, pair B, pair C) { return(Circle(mixtilinearcirclecenter(A,B,C),mixtilinearcircleradius(A,B,C))); }; pair mixtilinearcircletangentpoint(pair A, pair B, pair C) { pair[] R; R[0] = abs(circumcenter(A,B,C) - A); R[1] = mixtilinearcircleradius(A,B,C); return((R[0]*mixtilinearcirclecenter(A,B,C) - R[1]*circumcenter(A,B,C))/(R[0] - R[1])); }; unitsize(0.4 cm); pair A, B, C, Cp, D, E, F, G, O, P, Q, R; pair[] I; A = (0,0); B = (14,0); C = (2,12); I[0] = incenter(A,B,C); I[1] = excentre(A,B,C); I[2] = excentre(B,C,A); I[3] = excentre(C,A,B); D = (I[0] + I[3])/2; E = (I[1] + I[2])/2; P = mixtilinearcircletangentpoint(C,A,B); Q = intersectionpoint(I[0]--E,mixtilinearcircle(C,A,B)); O = circumcenter(A,B,C); Cp = intersectionpoint(interp(P,C,0.1)--C,mixtilinearcircle(C,A,B)); draw(mixtilinearcircle(C,A,B),red); draw(E--C--D); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(C--P--E); draw(C--Q--Cp); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NW); dot("$C'$", Cp, SE); dot("$D$", D, S); dot("$E$", E, N); dot("$P$", P, SW); dot("$Q$", Q, SE); dot(mixtilinearcirclecenter(C,A,B)); [/asy][/asy]

Dear Mathlinkers, you can see : http://jl.ayme.pagesperso-orange.fr/Docs/Mixtilinear1.pdf p. 23-26 Sincerely Jean-Louis

v_Enhance wrote: Let $\ell$ denote the angle bisector of $C$. Let $A_1$ and $B_1$ be the reflections of $A$ and $B$ in $\ell$, respectively. Set $Q_1 = CQ \cap AB$. We wish to show $CP$ and $CQ_1$ are isogonal. Note that by homothety, $Q_1$ is the tangency point of the $C$-excircle on side $AB$ of $\triangle ABC$. Consider an inversion which sends $A$ to $B_1$ and $B$ to $A_1$. Then $\omega$ is sent to the $C$-excircle of $\triangle CA_1B_1$ and $\Omega$ is sent to $A_1B_1$. So $P^\ast$ is the tangency point of the $C$-excircle of $\triangle A_1B_1C$ to side $A_1B_1$. In other words, $P^\ast$ is the reflection of $Q_1$ over $\ell$. Therefore, $CP$ is isogonal to $CQ_1$ as desired. This solution is awesome...

Dear MLs, M.T.

Let $\omega',$$R,$ $X$ be the incircle of $\triangle ABC,$ the touch-point of $C-\text{excircle}$ with $AB$ and the center of positive homothety $\omega'\mapsto \Omega$ respectively. By the homothety wrt $C$ and three homothety centers theorem $R\in CQ,$ $X\in AP$. Hence $$\angle ACP=\angle ACX=\angle RCB=\angle QCB.$$

$\omega$ is a mixtilinear circle. Let $I$ be incenter of $ABC$ and $T$ be midpoint of arc $ACB$ and $T'$ be midpoint of $AB$. It's known that $P,I,T$ are collinear. Let $PC$ meet $\omega$ at $C'$. Note that we just need to prove $\angle C'CT' = \angle QCT'$. Note that $CI$ passes through $O'$ center of $\omega$ so we need to prove $C'Q \perp CI$. Note that $\Omega$ and $\omega$ are tangent and tangent at $T$ to $\Omega$ is parallel with $AB$ so $P,Q,T$ are collinear so $C'Q || CT$ which $CT \perp CT'$. we're Done.

Invert about the circle centered at $C$ with twice the radius of $\Omega$. Since $\triangle ABC\sim \triangle A^*B^*C$ and $P^*$ is the $C$-excircle touchpoint of $\triangle A^*B^*C$ (since $\Omega$ maps to $A^*B^*$), we are immediately done. $\blacksquare$

We are trying to show that $CQ$ and $CP$ are isogonal. Note that if $CQ$ is extended to hit $AB$, it will hit it at the excircle touchpoint by a homothety around $C$. Perform a $\sqrt{ab}$ inversion about $C$. For any point $S$, let $S*$ denote its image. Then, $\omega$ gets sent to the C-excircle. Furthemore, $\Omega$ just becomes line $A*B*$, and since they remain tangent, $P*$ is the excircle touchpoint in triangle $\triangle CA*B*$. Thus, this implies $CQ$ and $CP$ are isogonal and we are done.

Does anyone know how to proceed? EDIT: nvmd my diagram was inconsistent with the problem but the solution's still there

Rewrite the problem to be $A$-indexed. A homothety sending $\omega$ to the $A$-excircle also sends $Q$ to $D$, the $A$-extouch point. I claim that $P$ and $D$ are swapped under $\sqrt{bc}$ inversion. Indeed, this inversion swaps $\overline{BC}$ and $(ABC)$, so it is clear that $\omega$ gets sent to some circle on the other side of $\overline{BC}$ as $A$ tangent to $\overline{AB},\overline{AC},\overline{BC}$, i.e. the $A$-excircle, and $P$ gets sent to $D$ as desired. $\blacksquare$

$A$-index the problem, and let $E$ be the excentral touchpoint, which obviously lies on $\overline{AQ}$. Consider the $\sqrt{bc}$ inversion that sends $\omega$ to the $A$-excircle. As this swaps $\overline{BC}$ and $(ABC)$, $P$ and $E$ are swapped, implying desired isogonality.

This is basically EGMO lemma 4.40. Let the line parallel to $AB$ and tangent to $\omega$ intersect with $AC,BC$ at $X1,Y1$ and let the touch points of $AC,BC$ with $\omega$ be $X,Y$. $CQ\cap AB=Q1$. Note that the incenters and excenters of $\triangle CX1Y1$ and $\triangle ABC$ are collinear,and from similarity we can easily get that $Q1$ is,indeed,the touch point of $C-excircle$ in $\triangle ABC$.This concludes our solution.

Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$. Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$. Thus $TI$ and $TA$ are a median and symmedian, respectively, of $\triangle TKL$, so $\angle KTA = \angle ITL = \frac C2$. As a result, it's clear that $TI$ will pass through the top point of $(ABC)$, and hence will also pass through the top point of the mixtilinear incircle. Suppose $TA$ and $TI$ meet the mixtilinear incircle at $E$ and $F$. Then $\triangle AKE \cong \triangle ALF$, giving the desired. $\blacksquare$

The problem is a little bit strange in it's direction, let's spin it. 2013 EGMO P5 wrote: Let $\Omega$ be the circumcircle of the triangle $ABC$. The circle $\omega$ is tangent to the sides $AB$ and $AC$, and it is internally tangent to the circle $\Omega$ at the point $P$. A line parallel to $BC$ intersecting the interior of triangle $ABC$ is tangent to $\omega$ at $Q$. Prove that $\angle PAB = \angle QAC$. Now let's get to the solving. Invert wrt $A$, with radius $\sqrt{bc}$, followed by a reflection through $AI$. Let the point $X$ be sent to $X'$. Now the mixtilinear incircle become the $A$-excircle. So $P'$ would be the foot from $I_A$ to $BC$. So $AP'$ passes through $Na$, which means $AP$ passes the triangle center $X_{56}$, or $Na^*$. Let the tangent of $\omega$ parallel to $BC$ intersects $AB, AC$ at $F, E$ respectively. Now consider a homothety centered at $A$, sending $F$ to $B$ and $E$ to $C$. We know that $\omega$ is the $A$-excircle of $AEF$, so $Q$ would be sent to $P'$. This means $AQ$ passes thorugh $Na$, with $AP$ passes through $Na^*$, which is the isogonal conjugate of $Na$, we're done.

Consider the $C$-excircle meeting $\overline{AB}$ at point $Q'$. Now observe that $C, Q, Q'$ are collinear due to the homothety at $C$ sending the line tangent to $\omega$ to $\overline{AB}$. Consider now an inversion $\Psi$, at $C$ with radius $\sqrt{ab}$ followed by a reflection over the angle bisector of $\angle ACB$. Notice that the image of the $C$-excircle remains a circle under $\Psi$ and simultaneously remains tangent to $\overline{CA}$ and $\overline{CB}$. Also the image of this excircle meets $\Psi(AB)$ at one point. Hence $\Psi(Q') = P$, as $\Psi(Q')$ must lie on a circle tangent to both $\overline{CA}, \overline{CB}$ and to $(ABC)$. Therefore \[\angle ACP = \angle Q'CB = \angle QCB, \]as desired. $\blacksquare$

Shift $\triangle ABC \to \triangle CBA$. Let $\Gamma$ be the $A$-excircle. By a homothety sending $\Gamma \to \omega$ we get that $\angle BAQ = BAD$ where $BC \cap \Gamma$. Then apply force-overlaid inversion on $\triangle ABC$. Since $\Gamma$ is tangent to $BC$, $\Gamma^\ast$ is tangent to $BC^\ast = \Omega$ and $\Gamma^\ast$ is tangent to $AB$ and $AC \implies \Gamma^\ast = \omega$, which finishes as $D^\ast = P$.

Well well well. Proof sketch Step 1: rtbc invert. Step 2: homothety at A

We wish to show that $CP$ and $CQ$ are isogonal, or that they are mapped to each other under reflection over the angle bisector. Let the tangents of $\omega$ to $BC$ and $AC$ be $X$ and $Y$, respectively. Let the circle centered at $C$ with radius $CX=CY$, denoted $\Gamma$. Consider the inversion around $\Gamma$, where $A^*$ denotes the image of $A$. $\omega$ is fixed, due to its tangency to $AC$ and $BC$ making it orthogonal to $\Gamma$. Then, $\Omega$ is sent to a line $A^*B^*$ which is antiparallel to $AB$ with respect to $AC$ and $BC$. $A^*B^*$ is also still tangent to $\omega$ at $P^*$. After reflecting over the angle bisector, $A^*B^*$ is then parallel to $AB$ and tangent to $\omega$, which is the definition of $Q$, so inverting back finishes.

Firstly, we can prove this problem for 2 steps; $1)$ Let $D$ be the point that $C-excircle$ is tangent to $AB$. Then prove $C,Q,D$ are collinear (It's done with homothety) $2)$ Then prove that $CD$, $CP$ are isogonal conjugates -Inversion from $C$ with radius $\sqrt{CA*CB}$ around $C$ angle bisector. Let that map be $\Psi$. Then it swaps $C-excircle$ with the $C-mixtillinear$ $incircle$. From there we get; $\angle BCQ=$$\angle ACP$ ==> they are isogonal conjugates, so we are done.