Prove that $(A,N;P,B)=(A,M;Q,C)=-1$ . Given $\Delta ABC$ and incircle $\omega$ tangent to $BC,AC,AB$ at $Y,M,N$ respectively . Let $AY \cap \omega=X$ . Let the tangent through $X$ wrt $\omega$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively .
Prove that $(A,N;P,B)=(A,M;Q,C)=-1$ .

I am completely stuck in this problem. I tried considering the intersection of $FE$ and $BC$, but no movement .
 A: If you familiar with harmonic quadrilaterals, then this can be proved in the following way.
Let $T$ be the intersection of the incircle and line $YP$ (and $Y\neq T$). Then, the quadrilateral $YNTX$ is harmonic. Indeed, it's cyclic and tangents at points $X$ and $Y$ intersect at point $P$ on line $YT$. Thus,
$$
(Y,T;N,X)=-1.
$$
Now, note that
$$
(Y,T;N,X)=(B,P;N,A),
$$
because we can project the incircle onto line $AB$ from point $Y$ (lines $YY$, $YT$, $YN$ and $YX$ intersect line $AB$ at points $B$, $P$, $N$ and $A$, respectively). Therefore,
$$
(B,P;N,A)=-1,~\text{so}~(A,N;B,P)=-1,
$$
as desired. Similarly, we can show that $(A,M;Q,C)=-1$.
A: It's well known that $BQ, CP, XY, NM$ concur say at $O$. Taking perspectivity at $O$, we get
$$(AM;QC)\overset{O}{=}(AN;PB)$$
To show that this is a harmonic bundle note that $$(AN;PB)=-1\implies \frac{AP}{NP}=\frac{AB}{NB}\iff \frac{AP}{PX}=\frac{AB}{BY}$$
Now as $\angle PXY =\angle BYX$, we get $\frac{AP}{PX}=\frac{AB}{BY}$ by ratio lemma. Completeing the proof.
A: Let $Z$ be a point on $AY$, different from $X$, such that $PX=PZ$. Then $\angle XZP = \angle PXZ = \angle XYB$, hence $PZ \parallel BY$. Hence $\dfrac{AP}{PZ} = \dfrac{AB}{BY}$. Since $PZ=PX=PN$ and $BY=BN$, we obtain $\dfrac{AP}{PN} = \dfrac{AB}{BN}$ which is equivalent to $(A,N;P,B)=-1$.
