The problem follows nicely by a "lemma" proposed by Virgil Nicula here: http://www.mathlinks.ro/Forum/viewtopic.php?t=132338 Now, returning to the problem: Let the incircles of $\triangle{ABX}$ and $\triangle{ACX}$ touch $BX$ at $D$ and $F$, respectively, and let them touch $AX$ at $E$ and $G$, respectively. Clearly, $DE \| FG$. If the line $PQ$ intersects $BX$ and $AX$ at $M$ and $N$, respectively, then $MD^{2}= MP\cdot MQ = MF^{2}$, i.e., $MD = MF$ and analogously $NE = NG$. It follows that $PQ \| DE$ and $FG$ and equidistant from them. The midpoints of $AB, AC$, and $AX$ lie on the same line $m$, parallel to $BC$. Applying the "lemma" to $\triangle{ABX}$, we conclude that $DE$ passes through the common point $U$ of $m$ and the bisector of $\angle{ABX}$. Analogously, $FG$ passes through the common point $V$ of $m$ and the bisector of $\angle{ACX}$. Therefore $PQ$ passes through the midpoint $W$ of the line segment $UV$ . Since $U, V$ do not depend on $X$, neither does $W$.

Let the incircle of $\triangle{ABC}$ meet $BC,CA,AB$ at $D_1,E_1,F_1$ and have inradius $r$. Then setting $u=AE_1=AF_1$ and so on, we have $r^2=uvw/(u+v+w)$. Similarly, let $\triangle{ABX}$'s incircle meet $BX,XA,AB$ at $D_2,E_2,F_2$ and let $\triangle{ACX}$'s incircle meet $CX,XA,AC$ at $D_3,E_3,F_3$. By simple length chasing, we find that $D_2D_3=E_2E_3=CD_1=CE_1=w$. Now define $M=BX\cap PQ$ and $N=AX\cap PQ$. Then $MD_2^2=MQ\cdot MP=MD_3^2$, and so $MD_2=MD_3=NE_2=NE_3=w/2$. Clearly we also have $D_3E_3 \| MN \| D_2E_2$, so \[\angle{NMX}=\frac\pi2-\frac{\angle{AXB}}{2}=\frac{\angle{D_1BF_1}}{2}+\frac{\angle{E_2AF_2}}{2},\]and the tangent addition formula combined with some simple ratios and lengths gives us \begin{align*} \tan\angle{NMX}=\frac{\frac{r}{v}+\frac{s}{u+v-sv/r}}{1-\frac{r}{v}\frac{s}{u+v-sv/r}}=\frac{r(u+v)-sv+sv}{v(u+v-sv/r)-rs} =\frac{r^2(u+v)}{v(r(u+v)-sv)-r^2s} &=\frac{uvw(u+v)}{v(u+v+w)(r(u+v)-sv)-uvws}\\ &=\frac{uw(u+v)}{r(u+v)(u+v+w)-s(v+u)(v+w)}\\ &=\frac{uw}{r(u+v+w)-s(v+w)}. \end{align*}Also, \[BM=BD_2+MD_2=\frac{sv}{r}+\frac{w}{2}.\]Now consider the coordinate plane with $x$-axis $BM$ and origin $B=0$. The line $MPQN$ is \[y=\frac{uw}{r(u+v+w)-s(v+w)}\left(x-\frac{sv}{r}-\frac{w}{2}\right).\]But it's easy to check that the point \[\left(\frac{w}{2}+\frac{v(u+v+w)}{v+w},\frac{uvw}{r(v+w)}\right)\]always lies on this line, so we're done.

Let the incircle of $ABX$ be tangent to $BX$ at $B_1$ and $AX$ and $B_2$. Also, let the incircle of $ACX$ touch $CX$ at $C_1$ and $AX$ at $C_2$. Let $M_1$ be the midpoint of $B_1$ and $C_1$, and $M_2$ be the midpoint of $B_2$ and $C_2$. Since $PQ$ is a radical axis, it passes through $M_1$ and $M_2$. Thus, it is sufficient to prove that, as $X$ varies, there is a constant point which $M_1M_2$ always passes through. Let $D$, $E$, and $F$ be the midpoints of $AB$, $AC$, and $AX$ respectively. Since $B$, $C$, and $X$ are collinear, the points $D$, $E$, and $F$ are collinear also. I will prove that $K=DE\cap M_1M_2$ does not change as $X$ changes, this will show that all $M_1M_2$ pass through a common point. It is sufficient to prove that $DK$ is a constant. Since $B_1,B_2, C_1,$ and $C_2$ are points of tangency of incircles, $B_1X=B_2X=\frac{BX+AX-AB}{2}$ and $C_1X=C_2X=\frac{CX+AX-AC}{2}$. Thus, \[M_1X=M_2X=\frac{BX+CX+2AX-AC-AB}{4}\]Since $DF\|BX$, $M_2FK\sim M_2XM_1$. Since $M_1X=M_2X$, $M_2F=KF$. We can now compute the length of $DK$ (notice the signed lengths): \begin{align*} DK&=DF-KF\\ &=\frac{BX}{2}-M_2F\\ &=\frac{BX}{2}-M_2X+FX\\ &=\frac{BX}{2}-\frac{BX+CX+2AX-AC-AB}{4}+\frac{AX}{2}\\ &=\frac{BX-CX+AC+AB}{4}\\ &=\frac{BC+AC+AB}{4} \end{align*} This is independent of $X$, so all $M_1M_2$ (and so $PQ$) intersect at a common point, as desired.

another solution : Let , U is exenter of triangle ABC opposite to A , I is incenter of ACX , incircle of CXA is tangent to XB at point T , incircle of ABX is tangent to BX at point K , incircle of BCA is tangent to BC at point L Easy to see that KT = LC A - Excircle of ABC is tangent to XB at point H Let line thru L' = midpont of LC and || to CI intersect line thru H' midpoint of BH and perpendicular to BU at point D , let line thru point D and perpendicular to XI intersect XB at point N Easy to see that angle NDL' = IAC , angle L'DH' = CAU , angle DH'N = AUI , so H'L'ND ~ UCIA , UC/CI = H'L'/L'N = HC/CT = H'L'/CT , L'N = CT , so N is midpoint of KT , line ND = line PQ , D is fixed . done

Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABX$ and $\triangle ACX$, respectively. Denote by $(P, \omega)$ the power of $P$ with respect to circle $\omega$. Define a function $f: \mathbb{R}^2 \to \mathbb{R}$ by\[f(P) = (P, \omega_1) - (P, \omega_2).\]This function is linear. We now use barycentrics with respect to $\triangle ABC$. Let $R = (x:y:z)$ be the constant point that lies on $PQ$. Since $R$ lies on the radical axis of $\omega_1$ and $\omega_2$, we have $f(R) = 0$. Let $BC = a$, $CA = b$, $AB = c$, $AX = p$, and $CX = q$. We claim that the point $R = (2a: a-b-c: a+b+c)$ works, which is independent of the position of $X$. We can compute \begin{align*} f(A) &= \left(\frac{a+c+p+q}{2}-a-q\right)^2 - \left(\frac{b+p+q}{2}-q\right)^2 \\&= \frac{1}{4}(-a+b+c+2p-2q)(-a-b+c), \\ f(B) &= \left(\frac{a+c+p+q}{2}-p\right)^2 - \left(\frac{b+p+q}{2}-p+a\right)^2 \\&= \frac{1}{4}(3a+b+c-2p+2q)(-a-b+c), \\ f(C) &= \left(\frac{a+c+p+q}{2}-c-q\right)^2 - \left(\frac{b+p+q}{2}-p\right)^2 \\&= \frac{1}{4}(a+b-c)(a-b-c+2p-2q). \end{align*} By linearity, we have $f(R) = xf(A)+yf(B)+zf(C)$. But now it's straightforward to check that indeed $2af(A)+(a-b-c)f(B)+(a+b+c)f(C) = 0$, implying that $R$ always lies on the radical axis of $\omega_1$ and $\omega_2$, as desired.

Short Solution: Take $B'$ and $C'$ fixed points that lie on $MN$ (line connecting midpoints of $AB$ and $AC$) such that $AB'B = AC'C = 90$. Let $K$ be the midpoint of $B'C'$. I claim $K$ is the desired point. First I prove a lemma LEMMA In triangle $XYZ$, let $X'$ and $Z'$ be the points of tangency of the incircle (centered at $I$) with $YX$ and $YZ$ respectively and let $W$ be the intersection of the bisector of angle $Z$ and $X'Z'$. Then, if $M$ and $N$ are the midpoints of $XY$ and $XZ$, we have $MNW$ are collinear and $XWZ=90$. Proof: Angle chasing Therefore, $B'$ lies on the line of the points if tangency of the incircle of $ABC$ with $AX$ and $BC$, and analogously $C'$. From this we see clearly $K$ lies on the radical axis, and we're done.

Let $\omega$ be the circle centered at the midpoint of the two incircles with radius the average of the other two. Let $Y$ be the point on $BC$ such that $\omega$ is the incircle of $AYX$. Note that \[ AY+AX-YX=\frac{AX+AC-CX}{2}+\frac{AX+AB-BX}{2} \] so \[ 2AY+BY-CY=AB+AC \] There are finitely many points $Y$ on $BC$ satisfying this, and $Y$ as a function of $X$ is continuous, so $Y$ must be the same for all $X$. Note that $PQ$ meets $XY,XA$ at their tangency points with $\omega$, so it suffices to show that for all $X$ varying on ray $BY$ past $Y$, there exists a constant point on all $X$-touch chords of $AYX$. But by a well known lemma, the intersection of the bisector of $AYX$ and the circle with diameter $AY$ suffices.

Let the $A$-midline of $\triangle ABC$ intersect the internal bisector of $\angle B$ at $V$ and the external bisector of $\angle C$ at $W$. Let $L$ be the midpoint of $\overline{VW}.$ We will show that $L \in \overline{PQ}$ to prove the result. Note that $\overline{PQ}$ is the mid-parallel of the $X$-touch chord of the incircles of $\triangle ABX$ and $\triangle ACX$. Point $V, W$ lie on the $X$ touch chord of $\triangle ABX$ and $\triangle ACX$, respectively (from the "right angles on the intouch chord" lemma). Evidently, $L$ being the midpoint of $\overline{VW}$ lies on $\overline{PQ}$ as desired.

Lemma:In $\triangle ABC$ the $B$-bisector $A$-midline and the $C$-touch chord of the incircle are concurrent Proof: Take the polar dual and it suffices to show that the line connecting $C$ and orthocenter of $\triangle CIB$ is perpendicular to $AI$ which is immediate. Let the incircles of $\triangle ABX,\triangle $ touch AX,BC in $E,F,E_1,F_1$.Now let $EF,E_1F_1\cap \text{A-midline}=\{R,S\}$.By lemma $R$ lies on the B-angle bisector and $S$ on C- outer angle bisector and hence they're fixed.$PQ||EF||E_1F_1$ and $PQ$ bisects $EE_1$ and hence it bisects $RS$ $\implies$ it passes thru the midpoint of $RS$ which is fixed.$\blacksquare$

Really nice problem! Let the incircle of $\Delta ABX, \Delta ACX$ centered at $I_1, I_2$ and touches $BC$ at $E, F$ resp. Let $I_B$ be the center of $B$-excircle of $\Delta ABC$, which touches $BC$ at $T$. Let $K, M$ be the midpoint of $AT, EF$ and finally let $R$ be the point on $BC$ such that $AR\perp I_1I_2$. Note that $M$ lies on radical axis of those incircles, which is $PQ$. Now we claim that $K$ also lies on $PQ$, which is the fixed point. It suffices to show that $MK\perp I_1I_2$, which is parallel to $AR$. Or equivalently, $M$ is the midpoint of $RT$. To prove this, note by symmetry that $XR=XA$. So by side-chasing, \begin{align*} ET &= CT - CF \\ &= \frac{AB + AC - BC}{2} - \frac{AC + CX - AX}{2} \\ &= \frac{AB + AX - BC - CX}{2} \\ &= \frac{AB + AX - BX}{2} \\ \end{align*}and \begin{align*}FR &= XR - EX \\ &= XA - \frac{XA + XB - AB}{2} \\ &= \frac{XA + AB - XB}{2} \\ \end{align*}so $ET=FR$ which implies $M$ is midpoint of $RT$ and we are done. Note : The most effective way to claim the fixed point is to plug $X=C$ and $X={\infty}_{BC}$ and intersect the radical axii.

Nice algebra problem! Solved with eisirrational. We use barycentric coordinates on $\triangle ABC$. Let $CX = d$ and $AX = x$. Moreover, let the incircles of $ABX$ and $ACX$ be tangent to line $AX$ at $M_1$, $N_1$, and to line $BX$ at $M_2$, $N_2$, respectively. I claim that the point $R = \left( \frac{1}{2}, \frac{a-b-c}{4a}, \frac{a+b+c}{4a} \right)$ is the desired point. It suffices to show that $R$ lies on the radical axis of the two incircles, i.e. $R$, the midpoint of $\overline{M_1N_1}$, and the midpoint of $\overline{M_2N_2}$ are collinear. Call these midpoints $S_1$ and $S_2$. We first compute some lengths. By equal tangents, we have that $N_1X = N_2X = \frac{x+d-b}{2}$ and $M_1X = M_2X = \frac{a+d+x-c}{2}$, so $S_1X = S_2X = \frac{2x+2d+a-b-c}{4}$. The coordinates of $S_2$ is readily calculated to be (with directed ratios) $$(0:CS_2:BS_2) = (0:CX-XS_2:BX-XS_2) = \left(0,\frac{a-b-c+2x-2d}{4a},\frac{3a+b+c-2x+2d}{4a} \right).$$The coordinates of $S_1$ are a bit harder, but still not too bad. Note the coordinates of $X$ are $(0:CX:BX) = \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right)$, so $$S_1 = XS_1 \cdot (1,0,0) + AS_1 \cdot \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right) = \left( \frac{2x+2d+a-b-c}{4x}, \frac{-2xd+2d^2+ad-bd-cd}{4ax}, \frac{2ax+2dx-3ad-d^2-a^2+ab+bd+ac+cd}{4ax}\right).$$To show that $X,S_1,S_2$ are collinear, we simply show that the displacement vectors $\overrightarrow{XS_1}$ and $\overrightarrow{XS_2}$ are proportional. In fact, because all of these coordinates are homogenized, we only to verify the proportion for the first two coordinates. Looking at the first coordinates, we receive a ratio of $$\frac{\frac{1}{2} - \frac{2x+2d+a-b-c}{4x}}{\frac{1}{2}} = \frac{-2d-a+b+c}{2x}.$$But for the second coordinates we also have $$\frac{\frac{a-b-c}{4a} - \frac{2xd+2d^2+ad-bd-cd}{4ax}}{\frac{a-b-c}{4a} - \frac{a-b-c+2x-2d}{4a}} = \frac{(d-x)(2d+a-b-c)/4ax}{(2x-2d)/4a} = \frac{-2d-a+b+c}{2x}.$$We thus have the desired.

Let $I_1$ be the incenter of $ABX$ and let $I_2$ be the incenter of $ACX$. Let $D$ and $F$ be the contact points of the incircle of $ABX$ with $BX$ and $AX$ respectively, and let $E$ and $G$ be the contact of the incircle of $ACX$ with $CX$ and $AX$ respectively. Let $P$ and $Q$ be the intersections of $DF$ and $EG$ with the $A$-midline respectively. The key claim is that $P$ and $Q$ are fixed as $X$ varies. This follows from the so called Iran lemma, as $Q$ is the intersection of the internal $C$-angle bisector with the midline and $Q$ is the intersection of the external $B$-angle bisector with the midline. Note that the midpoints of $DE$ and $FG$ are on the radical axis of the two circles, so the intersection of the radical axis with $PQ$ is simply the midpoint of $PQ$, which is a fixed point, as desired.

Let the incircle of $\triangle ABX$ touch $XA$ and $XB$ at $S$ and $T$, and let the incircle of $\triangle ACX$ touch $XA$ and $XC$ at $U$ and $V$. Let the incenters of $\triangle ABX$ and $\triangle ACX$ be $I$ and $J$ respectively, and let $Y$ and $Z$ be the foots of perpendicular from $A$ to $\overline{BI}$ and $\overline{CJ}$. Let $M$ be the midpoint of segment $YZ$. We claim that $M$ is the desired fixed point. By the Iran lemma, it follows that $Y$ and $Z$ lie on $\overline{UV}$ and $\overline{ST}$ respectively. It is also easy to see that $UV\parallel ST$ since both of them are perpendicular to $\overline{IX}$. Moreover, the radical axis passes through the midpoints of segments $US$ and $VT$, so it is actually the midline of $\overline{UV}$ and $\overline{ST}$, and this passes through $M$ as desired.

First ISL G7. Here's solution using famous Iran Lemma: Let $I_{1}$ and $I_{2}$ be the incenters of $\triangle{ABX}$ and $\triangle{ACX}$, respectively. Denote by $T_{1}$, $T_{2}$ touchpoints of incircle of $\triangle{ABX}$ with $\overline{BX}$ and $\overline{AX}$, respectively. Similarly, let incircle of $\triangle{ACX}$ touches $\overline{CX}$ and $\overline{AX}$ at $T_{3}$ and $T_{4}$, respectively. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AX}$, respectively. By Iran Lemma we have $T = T_{1}T_{2} \cup BI_{1} \cup MN$ and $S = T_{3}T_{4} \cup CI_{2} \cup MN$ are fixed, since $MN$, $BI_{1}$, and $CI_{2}$ are fixed.

gives us $T_2T_1T_3T_4$ is cyclic. From angle chasing we also get, that $T_1T_2 \parallel T_3T_4$. Hence, $TT_1T_3S$ is parallelogram. It's well-known, that $PQ$ bisects $\overline{T_1T_3}$; since $TT_1T_3S$ is parallelogram , it's also pass through the midpoint of $\overline{TS}$, which is fixed. $\blacksquare$

Let incircles of $ABX,ACX$ meet $AX$ again at $D_1,D_2$ and $BX$ at $E_1,E_2$ respectively. Let also line homothetic to $BC$ wrt $A$ with coefficient $\frac{1}{2}$ intersect internal bisector of and $ABC$ and external bisector of $ACB$ at $X,Y$ respectively, so by Iran lemma $X\in D_1E_1,Y\in D_2E_2.$ But obviously $D_1XD_2Y$ is an isosceles trapezoid, therefore $PQ$ passes through midpoint of $XY,$ which is fixed.

Let $M,N$ be midpoints of $AB,AC$. Incircle of triangle $ABX$, which has center $I$ touches $XA,XB$ at $K,L$. Let incircle of triangle $ACX$ touches $AX,CX$ at $J,G$. Let $D=PQ\cap MN, E=PQ\cap BC, F=BI\cap MN$ and $R=PQ\cap AX$. From Iran Lemma $F\in KL$. Since $RK=RJ$ and $El=EG$, we get $KL||RE \implies FDEL$ is parallelogram $\implies FD=EL=\frac{LG}{2}=\frac{XL-XG}{2}=\frac{(XA+XB-AB)-(XC+XA-AC)}{4}=\frac{BC+AC-AB}{4}$. Also $MF=MB=\frac{AB}{2} \implies MD=MF+FD=\frac{(BC+AC-AB)+2AB}{4}=\frac{AB+BC+CA}{4}$, which is constant when $ABC$ is fixed. So $D$ lies on fixed line $MN$ and length of $MD$ is constant, which implies $D$ is fixed point as $X$ varies. So we are done.

Let the incircle of $ABX$ intersect $BX$ at $D$ and $AX$ at $E$, and let $M$, $N$ be the midpoints of $AB$ and $AX$ respectively. Let $I$ be the incenter, then we claim that $BI$, $MN$, $DE$ are concurrent. Let $S$ be the intersection point of $DE$ and $MN$. Let $F$ be on $DE$ such that $AF\parallel BX$. Let $G$ be $AS$ intersection with $BX$. Let $R$ be the contact point on $AB$. Clearly, $AFGD$ is a parallelogram with center $S$. We have \[\angle AEF = \angle DEX = \angle EDX = \angle AFE\]so $AE=AF$. Thus, $DG=AF=AE=AR$. Also, $BD=BR$ so $AB=AG$. Since $AS=SG$, $\triangle ABS\cong \triangle GBS$ so $S$ is on $AI$ as desired. Now, similarly, if $J$, $K$ are the points of contact of the incircle of $ACX$ on $CX$ and $AX$, respectively then $JK$ passes through $T$, a fixed point on the midline and the $C$-angle bisector. Let $U$ be the midpoint of $ST$, also a fixed point with respect to $ABC$. Since $XI\perp DE$ and $XI\perp JK$, $DE\parallel JK$. Now, since $PQ$ is the radical axes of the two circles, it bisects $EK$ and $DJ$. Thus, $PQ$ is the line right in the middle of $DE$ and $JK$, so it passes through $U$, as desired.

nice problem! glad i solved it in ~45 minutes? this is the last problem im doing on configgeo im done Define the points as in the diagram, with all the lines given by Iran Lemma; note that $YJ\perp CD\perp VW\implies YJ\parallel VW\parallel PQ$. On the other hand, since PQ is radax which bisects the common tangent RK, PQ is the midline of JYWV, which passes through the midpoint of YV. We conclude. also note that my diagram is extremely overkill i didnt need ALL of those info but its helpful to list out when you dont know what to do

Consider an arbitrary point $X$ on ray $BC$. Let $I$ be the incenter of $\triangle ABC$ with $\triangle DEF$ being the intouch triangle of $\triangle ABC$. Further, let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$ respectively. Let $M_{CD}$ and $M_{CE}$ be the midpoints of $CD$ and $CE$. We claim that all lines $PQ$ pass through $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$. Let $\omega$ be the incircle of $\triangle ABC$ and $\omega_B$ and $\omega_C$ the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Let $O_B$ and $O_C$ be the centers of $\omega_B$ and $\omega_C$ respectively. Now, let $G$ and $H$ be the tangency points of the incircle of $\triangle ABX$ with sides $AX$ and $BX$. Let $M$ be the midpoint of $AX$. Clearly, $B-I-O_B$ and $C-O_C-O_B$ due to the common tangents. Further, $M_C-M_B-M$. By Iran Lemma on $\triangle ABC$, we see that $\overline{BI},\overline{M_BM_C}$ and $\overline{DE}$ concur (say at $N$). Now, clearly $N$ also lies on $BI$ and $M_CM_B$ due to the above collinearities.Further notice that by Iran Lemma on $\triangle ABX$, we must have that lines $\overline{GH},\overline{M_CM}$ and $\overline{BO_B}$ concur. But clearly the latter two of these lines intersect at $N$. Thus, $GH$ also must pass through $N$. Now, simply notice that $\overline{ED}\parallel\overline{M_{CD}M_{CE}}$ by Midpoint Theorem and $\overline{GH} \parallel \overline{PQ}$ since both these lines are perpendicular to $XO_C$. Thus, the intersection of $EF$ and $GH$ and the intersection of $M_{CD}M_{CE}$ and $PQ$ must form two triangles which are similar. Now, we prove the following. Let $D',E'$ be the tangency points of $\omega_C$ with $AX$ and $BX$. Then, Claim : $GD'=CD$. Proof : Note that $XD'$ and $XG$ are tangents to $\omega_B$ and $\omega_C$ respectively. Then, let $s_1$ and $s_2$ denote that semiperimeters of $\triangle ABX$ and $\triangle ACX$ and $s$ denote the semiperimeter of $\triangle ABC$. We have, \begin{align*} GD' &= XG-XD'\\ &= s_1-AB - (s_2-AC)\\ &= s_1-s_2 + AC - AB\\ &= \frac{AX+XB + AB - AX - CX - AC}{2} + AC - AB\\ &= \frac{BC + AB - AC}{2} + AC - AB\\ &= \frac{AB+AC+BC}{2}-AB\\ &= s-AB\\ &= CD \end{align*}Thus, indeed $GD'=CD$ as claimed. Now, let $R = \overline{PQ} \cap \overline{BX}$. It is well known that the radical axis bisects the common tangent. Thus, $RD'=\frac{GD'}{2}=\frac{CD}{2}=DM_{CD}$ Thus, the previously described similar triangles must infact also be congurent. This means, that the intersections of $EF$ and $GH$ and of $M_{CD}M_{CE}$ and $PQ$ must lie on a line parallel to $BC$. But clearly the former intersection point is $N$ which lies on $\overline{M_CM_B}$ and thus $Z'= M_{CD}M_{CE} \cap PQ $ must also lie on $\overline{M_CM_B}$. This means, $Z'=Z$. Thus, all lines $PQ$ must pass through a common point, this point being $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$.

Define line $\ell$ as the $A$-midline and the touch points of the incircles of $\triangle ABX$ and $\triangle ACX$ to $AX$, $BX$ to be $J_1$, $J_2$ and $K_1$, $K_2$. If we suppose $J = J_1J_2 \cap \ell$ and $K = K_1K_2 \cap \ell$, we know $J$ and $K$ are fixed as they lie on the $B$-angle bisector and $C$-external angle bisector, respectively, through Iran Lemma. $PQ$ bisects both $J_1K_1$ and $J_2K_2$ by power of a point, and $J_1J_2 \parallel PQ \parallel K_1K_2$. Thus $PQ$ passes through the midpoint of $JK$, which is fixed. $\blacksquare$

I'm surprised no one found this clean linpop solution! Let $\omega_1$ and $\omega_2$ denote the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Define $f(\bullet) = \text{Pow}(\bullet, \omega_1) - \text{Pow}(\bullet, \omega_2)$. We claim that the intersection of the $A$-midline and the radical axis of $\omega_1$ and $\omega_2$ is fixed. In order to prove this, it suffices to show that the powers of $f((A+B)/2)$ and $f((A+C)/2)$ are independent of $X$, because it follows that the point on the $A$-midline that $f$ vanishes on is independent of $X$. This can be done by bashing. Here's an example of what the bash would look like; consider the example of proving $f((A+B)/2)=(f(A)+f(B))/2$ constant (proving $f((A+C)/2)$ is analogous). Let all lengths be signed. Then using the intouch points, we calculate \[ f(A) = \frac{1}{4} \left[ (AB+AX-BX)^2 - (AC+AX-CX)^2\right] \]and \[ f(B) = \frac{1}{4} \left[ (AB+BX-AX)^2 - (2BC+AC+CX-AX)^2\right]. \]Now we can bash out $f(A)+f(B)$, and utilizing the fact that $BC+CX=BX$, we compute this quantity to be independent of $X$, as desired.

This problem is so cute! I claim that the fixed point $K$ lies on the $A$-midline $\overline{MN}$. Let $E$ and $F$ are the tangency points of the incircle of $ABX$ to $\overline{BX}$ and $\overline{AX}$, respectively, and define $G$ and $H$ similarly. Let $I_1$ and $I_2$ be the incenters of triangles $ABX$ and $ACX$, and recall that $R = \overline{BI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{EF}$. Similarly, $S = \overline{CI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{GH}$. Note that $R$ and $S$ are fixed points. Now, by radical axis, $\overline{PQ}$ is the midline of trapezoid $EGHF$, i.e. $K = \overline{PQ} \cap \overline{MN}$ is the midpoint of $\overline{RS}$. It follows that $K$ is the desired fixed point.