Problem: Let $D$, $E$ and $F$ be the points of tangency between the incircle of triangle $ABC$ and the sides $BC, CA,$ and $AB$. Let $M$ and $N$ be the feet of the perpendiculars from $B$ onto $CI$, and from $C$ onto $BI$, respectively. Prove that the points $M, D, E,$ and $N$ are collinear.
So Far I have:
Since $BMDI$ is a cyclic quadrilateral this implies that $\angle BMI = 90^\circ = \angle BDI$.
We know that $\angle MDI + \angle MBI = 180^\circ$. Denote $\angle MDB = \theta$. Then $\angle MBI = 90^\circ - \theta$.
If we can only prove: $\angle MDI + \angle IDE = 180^\circ$, then we can prove $M, D, E$ are collinear. (and then use a similar method to prove $D, E, N$ collinear).
$\triangle BDI \equiv \triangle BFI$ by HL so $\angle DBI = 90^\circ - \theta - \beta = \angle FBI$ ($\beta = \angle MBD$) and $\angle DIB = \angle FIB = \theta + \beta$.
Any help on what I should do next is appreciated]1