Omid Hatami wrote: $I$ is incenter of triangle $ABC$. Incircle of $ABC$ touches $AB,AC$ at $X,Y$. $XI$ intersects incircle at $M$. Let $CM\cap AB=X'$. $L$ is a point on the segment $X'C$ that $X'L=CM$. Prove that $A,L,\boxed{Y}$ are collinear iff $AB=AC$. Maybe you made a bit of mistake : in the my opinion, I think ... $AB=AC\Longleftrightarrow$ the points $A,L,I$ are collinearly !

Omid Hatami wrote: $I$ is incenter of triangle $ABC$. Incircle of $ABC$ touches $AB,AC$ at $X,Y$. $XI$ intersects incircle at $M$. Let $CM\cap AB=X'$. $L$ is a point on the segment $X'C$ that $X'L=CM$. Prove that $A,L,I$ are collinear iff $AB=AC$. Proof. From a well-known property, $AX'=p-b\ .$ Therefore, $\boxed{\ L\in AI\ }\Longleftrightarrow$ $\frac{AX'}{AX}\cdot \frac{IX}{IM}\cdot\frac{LM}{LX'}=1$ $\Longleftrightarrow$ $\frac{LM}{LX'}=\frac{p-a}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{LX'}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{MC}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{CX'}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $\frac{2r}{h_{c}}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $2r(p-b+c)=2pr$ $\Longleftrightarrow$ $\boxed{\ b=c\ }\ .$

Virgil Nicula wrote: Omid Hatami wrote: $I$ is incenter of triangle $ABC$. Incircle of $ABC$ touches $AB,AC$ at $X,Y$. $XI$ intersects incircle at $M$. Let $CM\cap AB=X'$. $L$ is a point on the segment $X'C$ that $X'L=CM$. Prove that $A,L,I$ are collinear iff $AB=AC$. Proof. From a well-known property, $AX'=p-b\ .$ Therefore, $\boxed{\ L\in AI\ }\Longleftrightarrow$ $\frac{AX'}{AX}\cdot \frac{IX}{IM}\cdot\frac{LM}{LX'}=1$ $\Longleftrightarrow$ $\frac{LM}{LX'}=\frac{p-a}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{LX'}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{MC}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{CX'}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $\frac{2r}{h_{c}}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $2r(p-b+c)=2pr$ $\Longleftrightarrow$ $\boxed{\ b=c\ }\ .$ DEAR YOU, why $AX'=p-b\ .$ and $\frac{2r}{h_{c}}=\frac{c}{p-b+c}$ ?

See http://www.mccme.ru/~akopyan/papers/EnGeoFigures.pdf 2.8). Is well-know that $AX' = XB$.Since $X'L = MC$, $L$ is the Nagel point. So $A,L$ and $I$ are collinear if $AB = AC$ because $AI$ would altitude of $ABC$.