Prove $NP$ is parallel to $BC$

Question

Given triangle $\triangle ABC$, let $M$ be the midpoint of $BC$, $I$ the incenter of $\triangle ABC$, $L$ the foot of the external $A$ angle bissector on $BC$ ($AL \perp AI$).

Also, define:

$N = CI \cap AL$ and $P = AB \cap MI$. Prove that $NP \parallel BC$

this smells like projective geometry, I think the key is to assume lines $NP$ and $BC$ meet on $X$ and then show that $(B,C;M,X) = -1$ but I haven't found this projection yet. Projection (centered at $I$) of point $B$ on $NP$ is weird and I think $NP \cap AI$ doesn't help much. Any suggestions?

Very well observed: $N$ is the $C$ ex-incenter

Answer 1

I found the answer and I will not give points to answers after this one.

$R = NC \cap AB$, then $(C,R;I,Q) = -1$ (because $Q$ is the $C$-exincenter and we have $I$ and $Q$ as feet of the angle bissectors of $\triangle BCR$)

$-1 =(C,R;I,Q) \frac{P}{\overline \wedge} (C,B;M,PQ \cap BC)$

therefore $PQ \cap BC = \infty \iff PQ \parallel BC$

Answer 2

Sketch. $N$ is the $C$-excenter of the triangle $ABC$ as intersection of angle bissectors $AL$ and $CI$. Hence, $\angle IBN=90^{\circ}$. Now choose $P\in AB$ usch that $NP'\parallel BC$, then it is easy to check that $NP'=P'B$. Besides that, since $\angle IBN=90^{\circ}$ the line $BI$ intersects $NP'$ at point $R$ such that $P'$ is the midpoint of $NR$. Now $NRCB$ is a trapezoid, so the midpoints of $NR$ and $BC$ and $I=NC\cap BR$ are collinear. Thus, $P'$, $M$ and $I$ are collinear, so $P=P'$ and $NP\parallel BC$, as desired.