Bisector proof in an olympiad level problem I simplified this from a problem in IMO Shortlist:
The incircle Ω of the acute-angled triangle  is tangent to  at . Let  be an altitude of triangle  and let  be the midpoint of . If  is the other common point of Ω and , prove that NK is the bisector of angle BNC.

I have no idea on this. I tried by analytic and trigonometric way but the incircle radius is too complicated. Are there any euclidean tricks to solve it?
 A: As OP points out in the comments, the question arises from problem G7 in an IMO Short-listed Problems and Solutions document (also archived here). On pg 19, Solution 2 for that problem demonstrates that $NK$ bisects $\angle BNC.$
A: Here I'd like to provide two solutions: a synthetic one and a coordinate one.
Synthetic Solution
This solution is mentioned in a famous series of IMO training books in China, ISBN: 9787542850393, the Analytic Geometry part, provided by Libing Huang.
The solution uses some projective geometry knowledges and looks much simpler than two solutions in the original IMO 2002 shortlist G7. There are many additional points and lines added so some letters are changed:

D, E and F are tangent points of incircle (I) to edges BC, CA and AF respectively; AH is the altitude and M its the midpoint; D and N are intersections of (I) and DM. Prove ND bisects $\angle BNC$.

*

*Make $P=BC\cap EF$, such that P is on D's polar BC and A's polar EF, then according to this (Theorem 6), P is the pole of AD.


*Make Q the other intersection of (I) and PN, and make $R=PQ\cap AD$, then according to this (Theorem 7), $(P,N;R,Q)=-1$.


*Make $S=AH\cap QD$, then $(H,M;A,S)=(P,N;R,Q)=-1$ through perspective center D. Because $AM=MH$, S must a point at infinity, then $AH//QD$, then $PQ\perp ND$.


*Make $T=PE\cap AD$, then $(P,F;T,E)=-1$, then $(P,B;D,C)=(P,F;T,E)=-1$ through perspective center A. According to this (Problem 2), ND bisects $\angle BNC$.
Coordinate Solution
I just use the above diagram, then put D onto origin and BC onto x-axis. Then set $B=(-b,0)$, $C=(c,0)$ and $I=(0,r)$. Solve F by the fact that D and F are the two intersection of two circles:
$$\begin{cases}x^2+y^2-2ry=0\\x^2+2bx+y^2=0\end{cases}$$
Then solve E analogously, then get A, M and N, then their coordinates are:
$$F=\left(-\frac{2br^2}{b^2+r^2},\frac{2b^2r}{b^2+r^2}\right)\\E=\left(\frac{2cr^2}{c^2+r^2},\frac{2c^2r}{c^2+r^2}\right)\\A=\left(\frac{(b-c)r^2}{bc-r^2},\frac{2bcr}{bc-r^2}\right)\\M=\left(\frac{(b-c)r^2}{bc-r^2},\frac{bcr}{bc-r^2}\right)\\N=\left(\frac{2bc(b-c)r^2}{b^2c^2+b^2r^2-2bcr^2+c^2r^2},\frac{2b^2c^2r}{b^2c^2+b^2r^2-2bcr^2+c^2r^2}\right)$$
So it's easy to prove $BN:BD=CN:CD$.
