Points of intersection of pole lines with the sides of the triangle are collinear in $\mathbb{R}P^2$.

Consider the triangle $$\Delta PQR$$ in $$\mathbb{R}P^2$$ and a point $$S$$ outside the triangle. If $$l$$ is harmonically added to $$PS$$ with respect to $$\{PQ,PR\}$$, $$m$$ is harmonically added to $$QS$$ with respect to $$\{PR,QR\}$$ and $$n$$ is harmonically added to $$RS$$ with respect to $$\{PR,QR\}$$. Show that the intersection points $$l \cap QR, m \cap PR$$ and $$n \cap PQ$$ are collinear.

So my approach was that if we call the intersection points $$L= l \cap QR, M = m \cap PR$$ and $$N = n \cap PQ$$ that they are collinear if: $$(L,Q,R)(M,R,P)(N,P,Q) = 1$$ This is Menelaos' theorem. To get to this, I started with the transversality condition. If $$l,m,n$$ are the pole lines, one can consider for instance the $$4$$ lines: $$PQ,PR, PS$$ and $$PL$$. These are $$4$$ hyperplanes in the projective space with $$P$$ the intersection. If we consider the line $$QR$$, it intersects every hyperplane in one point: $$Q,R,S',L$$ where $$S'$$ is the intersection of $$QR \cap PS$$ and $$L$$ is the intersection of the pole line and $$QR$$. The transversality condition then says: $$(PQ,PR,PS,PL) = (Q,R,S',L)$$ Because the pole line is harmonically added, we know that $$-1 = (PQ,PR,PS,PL) = (Q,R,S',L)$$ If we do this anologous for the other pole lines, I find: $$(PQ,QR,QS,QM) = (P,R,S'',M) = -1$$ $$(PR,QR,RS,RN) = (P,Q,S''',N) = -1$$ With $$S'',S'''$$ other intersections analogous as $$S'$$. Because we have harmonic fournumbers, we can interchange the variables as: $$(S',L,Q,R) = -1$$ $$(S'',M,R,P) = -1$$ $$(S''',N,P,Q) = -1$$ I now make a connection between the cross-ratio and partial ratio: $$(S',L,Q,R) = \frac{(L,Q,R)}{S',Q,R}$$ $$(S'',M,R,P) = \frac{(M,R,P)}{S'',R,P}$$ $$(S''',N,P,Q) = \frac{(N,P,Q)}{S''',P,Q}$$

We then eventually find that: $$(L,Q,R)(M,R,P)(N,P,Q) = -1(S',Q,R)(S'',R,P)(S''',P,Q)$$ But now I'm stuck and don't now how to proceed. I feel like I'm close, but I maybe can have it totally wrong.

The answer was indeed very close, I missed one final element. To prove that: $$(S',Q,R)(S'',R,P)(S''',P,Q)=-1$$ is actually very easy. If we use Ceva's theorem, it says that if $$S'P \cap s''Q \cap S'''R \neq \emptyset$$, then: $$(S',Q,R)(S'',R,P)(S''',P,Q)=-1$$ And from our construction we have $$S'P \cap s''Q \cap S'''R = S$$ so we can use Ceva's theorem, so eventually we find: $$(L,Q,R)(M,R,P)(N,P,Q)=1$$ So $$L,M,N$$ are collinear.