*non-isosceles

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -13.954491762721387, xmax = 17.452195320260767, ymin = -8.930440031300787, ymax = 10.693607583948236; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((0,0), 6), linewidth(0.8) + red); draw((-5.031818084078979,3.2681503589547036)--(-5.164864287379107,-3.0535515212545397), linewidth(0.8) + blue); draw((-5.164864287379107,-3.0535515212545397)--(4.204835448909678,-4.280112013439899), linewidth(0.8) + blue); draw((4.204835448909678,-4.280112013439899)--(-5.031818084078979,3.2681503589547036), linewidth(0.8) + blue); draw(circle((-2.9807552223058917,-1.1770321463890603), 2.1441411359219087), linewidth(0.8) + red); draw((-5.124421661673871,-1.131916655923956)--(-1.6239732520539345,0.48323406598524243), linewidth(0.8) + blue); draw((-2.9807552223058917,-1.17703214638906)--(-5.031818084078979,3.2681503589547036), linewidth(0.8) + blue); draw((-5.031818084078979,3.2681503589547036)--(-3.2590640678998892,-3.3030343564576747), linewidth(0.8) + blue); draw((-5.031818084078979,3.2681503589547036)--(-1.1826005318809738,-2.3449280835578485), linewidth(0.8) + blue); draw((-4.29057351017264,0.5205300066770097)--(-1.1826005318809738,-2.3449280835578485), linewidth(0.8) + blue); draw((-3.2590640678998892,-3.3030343564576747)--(-3.422338371545886,0.9211445699672581), linewidth(0.8) + blue); draw(circle((-5.031818084078979,3.2681503589547036), 4.401041371994425), linewidth(0.8) + qqwuqq); draw((-2.9807552223058917,-1.17703214638906)--(-1.6239732520539345,0.48323406598524243), linewidth(0.8) + blue); draw((-2.9807552223058917,-1.17703214638906)--(-5.124421661673871,-1.131916655923956), linewidth(0.8) + blue); draw((-1.1826005318809738,-2.3449280835578485)--(-2.9807552223058917,-1.17703214638906), linewidth(0.8) + blue); draw((-4.29057351017264,0.5205300066770097)--(-2.9807552223058917,-1.17703214638906), linewidth(0.8) + blue); draw(circle((1.344809319213931,3.514621783820163), 6.38138897135265), linewidth(0.8) + red); /* dots and labels */ dot((-5.031818084078979,3.2681503589547036),dotstyle); label("$A$", (-4.9430305670160495,3.4680168302163783), NE * labelscalefactor); dot((-5.164864287379107,-3.0535515212545397),dotstyle); label("$B$", (-5.086721292232308,-2.8543750792989977), NE * labelscalefactor); dot((4.204835448909678,-4.280112013439899),linewidth(4pt) + dotstyle); label("$C$", (4.294230339743407,-4.106537113326394), NE * labelscalefactor); dot((-2.9807552223058917,-1.17703214638906),linewidth(4pt) + dotstyle); label("$I$", (-2.8903059210695035,-1.006922897947102), NE * labelscalefactor); dot((-3.2590640678998892,-3.3030343564576747),linewidth(4pt) + dotstyle); label("$A_1$", (-3.17768737150202,-3.1417565297315146), NE * labelscalefactor); dot((-1.6239732520539345,0.48323406598524243),linewidth(4pt) + dotstyle); label("$B_1$", (-1.5355076547447832,0.6557840652696039), NE * labelscalefactor); dot((-5.124421661673871,-1.131916655923956),linewidth(4pt) + dotstyle); label("$C_1$", (-5.045666799313377,-0.9658684050281712), NE * labelscalefactor); dot((-3.374197456863902,-0.32434129496935665),linewidth(4pt) + dotstyle); label("$M$", (-3.3008508502588128,-0.1653057931090164), NE * labelscalefactor); dot((-4.29057351017264,0.5205300066770097),linewidth(4pt) + dotstyle); label("$A_2$", (-4.204049694475293,0.6763113117290693), NE * labelscalefactor); dot((-3.422338371545886,0.9211445699672581),linewidth(4pt) + dotstyle); label("$P$", (-3.3419053431777437,1.0868562409183795), NE * labelscalefactor); dot((-1.1826005318809738,-2.3449280835578485),linewidth(4pt) + dotstyle); label("$Q$", (-1.1044354790960085,-2.1769759461366363), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice problem $\color{black}\rule{25cm}{1pt}$ We draw a circle centered at $A$ and radius $AB_1$. Because we have that $\angle AB_1I = 90$, this implies that the the constructed circle and the incircle are orthogonal. We shall use $\omega$ to denote the constructed circle. Let's call an inversion around $\omega$ with $\varphi_1$ and let's call an inversion around $\Gamma$ with $\varphi_2$. Since $\omega$ and $\Gamma$ are orthogonal this implies that $\omega \overset{\varphi_2}{\rightarrow} \omega$ and we have that $\Gamma \overset{\varphi_1}{\rightarrow} \Gamma$. We have that $M \overset{\varphi_1}{\rightarrow} I$, this is seen easily, because of the construction of point $M$. It is also easily seen that $M \overset{\varphi_2}{\rightarrow} A$, again this holds because of construction. Since we have orthogonal circles we have that $A_1 \overset{\varphi_1}{\rightarrow} A_2$. Let $P'$ be the point on the incircle such that $P \overset{\varphi_1}{\rightarrow} P'$. Since we have that $\angle AMP + \angle AMA_1 = 180$ this implies that: $$180=\angle AMP + \angle AMA_1 \overset{\varphi_1}{=} \angle AP'I + \angle AA_2I$$which in turn implies that $IA_2AP'$ is a cyclic quad. But on the other hand we have that: $$\angle IQA_2 = \angle IA_2Q = \angle IA_2M \overset{\varphi_2}{=} \angle IAA_2$$thus this implies that $IA_2AQ$ is a cyclic quad. Since we must have that $P'$ and $Q$ are on the incircle, this implies that $IA_2AQP'$ is cyclic., but notice that the circumcircle around $(IA_2A$ is fixed and $A_2$ is on the incircle, from here we come to the conclusion that $P' \equiv Q$, this implies that $A,P,Q$ are colinear.

llplp wrote: Since $A_1 A_2$ is a symmedian in $A_1 B_1 C_1$, the quadrilateral $A_1 B_1 A_2 C_1$ is harmonic. Projecting through $Q$ onto $B_1 C_1$ gives $$(Q A_1 \cap B_1 C_1, M ; B_1, C_1) = -1,$$or $Q A_1 \parallel B_1 C_1$, because $M$ is the midpoint of $B_1 C_1$. Now, projecting $(P,Q ; B_1, C_1)$ through $A_1$ onto $B_1 C_1$ perserves the harmonic ratio: $$(P,Q ; B_1,C_1) \stackrel{A_1}{=} (M, A_1 Q \cap B_1 C_1; B_1, C_1) = (M, \infty; B_1, C_1) = -1,$$therefore the quadrilateral $P B_1 Q C_1$ is harmonic aswell. This means $PQ$ is a symmedian in $Q B_1 C_1$ as desired.

Let $Q’=AP\cap \Gamma$. Notice that $A_1M$ and $A_1A_2$ are median and symmedian in $\triangle A_1B_1C_1$, so $A_2PB_1C_1$ is an isosceles trapezoid. But this means that $A_2$ and $P$ are symmetric about $AI$, and so $Q’$ and $D$ are symmetric about $AI$ too. Consequently $A_2Q’$, $PD$ and $AI$ are concurrent at the point $M$, so $Q’=Q$ and we’re done.