Let $I$ be the incenter of a triangle $\triangle ABC$. The circle $AIB$ meets the sides $BC$ and $AC$ at points $M$ and $N$, respectively. I'm trying to prove then that $BM=AN$. Here's a figure for reference:
Let $O$ be the center of the circle $AIB$. Obviously, I first tried showing that the triangles $\triangle BOM$ and $\triangle NOA$ are congruent to each other. In this direction, it's obvious that $OB=OM=ON=OA$. But I have no clue as to how to show $\angle BOM=\angle NOA$ !!
As that didn't work out, I started considering the cyclic quadrilateral $\square ANBM$. I am aware of many interesting properties of cyclic quadrilaterals (Ptolemy's Theorem, opposite angles being supplementary, diagonal intersection, etc.) but none of them seem to help towards actually being able to show that $BM=AN$.
Does anyone have any useful comments or hints for this? TIA.