Proving that $N,K,O$ are collinear. Consider the square with centre $O$ below: $X$ is a point chosen on $AD$ and a circle is inscribed in $\Delta ABX$. $N,K$ are the tangent points. Prove that $N,K,O$ are collinear.

My try:
I have chosen points $E,F$ on $AD,BC$ such that $AN=DE=CF$ and I created a rectangle $EFGN$ shown below. Obviously $O$ is the Centre of square and rectangle. It is enough to prove that $NK$ extended passes through $O$. I started with contradiction that let $NK$ does not pass through $O$, so let $NK$ extended meets $BC$ at $T \ne F$ and $EG$ at $R$.

Any help from here?
 A: 
Extend $NO$ to intersect $BC$ at $M$. Note that $NO$ might not pass through $K$ at this point. Connect $AC$, which passes through $O$. Label point $L$.
Since $CM=NA=AL$, we have $BM=BL=BK$. Also since $XN=XK$ and $\angle NXK=\angle KBM$ we have isosceles triangles $NXK$ and $KBM$ similar (connect $NK$ and $KM$ here) and the linearity of $N,K,M$ follows which implies linearity of $N,K,O$.
A: 
We can prove that $N, K, O $ are collinear by showing that $\angle NKB+ \angle BKO=180^o$
Proof:
Note that $O_1$ is the in-center of $\Delta ABX$.
We can assume that $\angle YBO_1= \angle KBO_1=\alpha$
$\therefore \angle AXB=90^o-2 \alpha$ and $\angle HXK=45^o-\alpha$
$\because \angle XHK=90^o$
$\therefore \angle NKB = 90^o+45^o-\alpha$
$\color{red}{ \angle NKB = 135^o-\alpha}$ ----- (1)
Next we consider $\angle BKO$.
$\because \angle O_1KB=90^o$ and $\angle O_1OB=90^o$
$\therefore \angle O_1KB = \angle O_1OB$
Hence $O, K, O_1, B$ are concyclic.
$\therefore  \angle BKO= \angle OO_1B$
But $\angle OO_1B=45^o+\alpha$
$\therefore \color{red}{\angle BKO=45^o+\alpha}$ ---- (2)
(1) and (2) $\implies $$\angle NKB+ \angle BKO=180^o$
Hence $N, K, O$ are collinear.
