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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.

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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.

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