[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 = -20.167733666790486, xmax = 28.928468233311115, ymin = -17.548208905647698, ymax = 13.128895026703361; /* image dimensions */ /* draw figures */ draw(circle((0,0), 6), linewidth(0.8) + red); draw(circle((-1.455213750217998,5.820855000871991), 4.7187646751672725), linewidth(0.8) + red); draw((-1.455213750217998,5.820855000871991)--(1.5828477936500027,-5.787451326977819), linewidth(0.8) + blue); draw((1.5828477936500027,-5.787451326977819)--(-5.214254425963497,2.9684256401870823), linewidth(0.8) + blue); draw((1.5828477936500027,-5.787451326977819)--(3.2039065451738695,5.072965882971424), linewidth(0.8) + blue); draw((-5.214254425963497,2.9684256401870823)--(3.2039065451738695,5.072965882971424), linewidth(0.8) + blue); draw(circle((-0.2604834159498832,1.2558402705577394), 2.862917800405046), linewidth(0.8) + red); draw(circle((1.5828477936500027,-5.787451326977819), 7.280510042311594), linewidth(0.8) + red); draw((-1.455213750217998,5.820855000871991)--(-5.164864287379107,-3.0535515212545397), linewidth(0.8) + blue); draw((-5.164864287379107,-3.0535515212545397)--(6,0), linewidth(0.8) + blue); draw((6,0)--(-1.455213750217998,5.820855000871991), linewidth(0.8) + blue); draw((-1.455213750217998,5.820855000871991)--(-5.214254425963497,2.9684256401870823), linewidth(0.8) + blue); draw((-1.455213750217998,5.820855000871991)--(3.2039065451738695,5.072965882971424), linewidth(0.8) + blue); draw((1.5828477936500027,-5.787451326977819)--(-5.164864287379107,-3.0535515212545397), linewidth(0.8) + blue); draw((1.5828477936500027,-5.787451326977819)--(6,0), linewidth(0.8) + blue); /* dots and labels */ dot((-1.455213750217998,5.820855000871991),dotstyle); label("$A$", (-1.3314784279933363,6.133488481460127), NE * labelscalefactor); dot((-5.164864287379107,-3.0535515212545397),dotstyle); label("$B$", (-5.021715825713391,-2.723081273068003), NE * labelscalefactor); dot((6,0),linewidth(4pt) + dotstyle); label("$C$", (6.113174409146253,0.2611976659577798), NE * labelscalefactor); dot((-0.2604834159498828,1.2558402705577394),linewidth(4pt) + dotstyle); label("$I$", (-0.14418465655297075,1.5126694790976243), NE * labelscalefactor); dot((3.2039065451738695,5.072965882971424),linewidth(4pt) + dotstyle); label("$M$", (3.321429595218907,5.331262960216637), NE * labelscalefactor); dot((-5.214254425963497,2.9684256401870823),linewidth(4pt) + dotstyle); label("$N$", (-5.08589386741287,3.213387584133823), NE * labelscalefactor); dot((1.5828477936500027,-5.787451326977819),linewidth(4pt) + dotstyle); label("$M'$", (1.7169785527319266,-5.514826086995348), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice and easy $\color{black}\rule{25cm}{1pt}$ Let $M'$ be the midpoint of the arc $BC$ that doesn't contain $A$. Then we have that: $$\angle AM'M=\angle ANM=\angle AMN = \angle AM'N$$By the reverse of the incenter-excenter lemma (or fact $5$) we have that $I$ is the incenter of triangle $NMM'$, since we have that $AN=AM$. Thus $MN$ is tangent to the incircle of $ABC$

parmenides51 wrote: Let $I$ be the incenter of a triangle $ABC$. The circle passing through $I$ and centered at $A$ meets the circumference of the triangle $ABC$ at points $M$ and $N$. Prove that the line $MN$ touches the incircle of the triangle $ABC$. (I. Kachan) Consider the invertion with center $A$ and radius $AM=AN$ then the incircle go to the mixtilinear. $T$ belongs to $(ABC),(E'F'T)$ so $X=T'$ belongs to $MN,(DEF)$ and because the two circle are tangent at $T$ we have the result

Attachments:

Let $K$ be the midpoint of minor arc $BC$. By the converse of incenter-excenter on $\triangle MKN$, $I$ is the incenter of $\triangle MKN$, which finishes.