We use barycentric coordinates. $A=(1,0,0),M=(0,\frac{1}{2},\frac{1}{2}),D=(0,\frac{s-b}{a},\frac{s-c}{a}),N=(\frac{1}{2},\frac{s-b}{2a},\frac{s-c}{2a}),I=(-a,b,c)$ Now we know that three points are collinear if and only if the value of the 3 by 3 determinant formed by their coordinates is equal to zero. Now the value of this determinant for $I,M,N$ is $(-a(\frac{1}{2}(\frac{s-c}{2a})-\frac{1}{2}(\frac{s-b}{2a}))+\frac{ba}{4}-\frac{ca}{4})$ which after simplification turns out to be zero.Hence $I,M,N$ are collinear.

see also: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=595536&p3533273#p3533273

Here's a quick way: let the incircle touch $BC$ at $E$ and let $DI$ intersect the $A$-excircle at $F$. It is well-known that $M$ is the midpoint of $ED$. But since the homothety bringing the incircle to the $A$-excircle brings $E$ to $F$, now $A,E,F$ are collinear, and so the midpoints of $AD,ED,FD$, ie. $N,M,I$, are collinear.

Why A,E,F collinear sir? Also what is Homothety?

Hi; a homothety is basically an enlargement, and it has some ratio of enlargement and a center. Let the incircle be $\omega$ and excircle be $\gamma$. To see that $A,E,F$ are collinear, look at points $E$ and $F$. If $I_a$ is the $A$-excenter, then note that $IE, I_aF$ are parallel, so $E,F$ must be corresponding points in the homothety sending $\omega$ to $\gamma$. This homothety must have center $A$, since $AB,AC$ are tangent to both circles. So the homothety centered at $A$ sends $E$ to $F$, and so $A,E,F$ are collinear.

Oh ok thank you

Can you give any question or problem more besides this one to study homothety?

Given circle $\omega$ and another circle $\omega '$ that is tangent to (at $Y$) and inside $\omega$, draw a line tangent to $\omega '$ at $X$ that cuts $\omega$ at $A$,$B$, show that $XY$ passes through the midpoint of arc $AB$ not containing $Y$.

it is my solution: AI cut BC at E , G is incenter of ABC. X,Y is midpoint of AD and BC. IE/IA=BE/BA=CE/CA=a/(b+c).XA/XD=1 ...YD=(b-c)/2,YE= a(b-c)/2(b+c) .menelauyt blah blah

Dear Mathlinkers, have a look at http://www.artofproblemsolving.com/community/c6t48f6h595536_prove_collinear Sincerely Jean-Louis