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$AL, BM, CN$ are proper Cevian lines and are concurrent at an ideal point.
To prove - Exactly one of the three points $L,M,N$ lies on the triangle $ABC$.
I was thinking that from Ceva's theorem we have, $\frac{BL}{LA} \cdot\frac{BN}{NC}\cdot\frac{BM}{MC} =1$. And $P$ is the only ideal point (I am thinking how?) and it is not in the ratio (which follows from the polarity)
It would help if you stated Ceva's theorem correctly. The correct statement is that $$\frac{BL}{LC}\cdot\frac{CM}{MA}\cdot\frac{AN}{NB} = 1$$ What does one of the points being on the triangle tell us? $\frac{BL}{LC}$ is positive if and only if $L$ is on segment $BC$, and similarly for the other two. In order for the product of all three ratios to be $1$, which is positive, either exactly one or all three of the points must be on the triangle.
Now, if two of the points $L$ and $M$ are on the triangle (sides $BC$ and $CA$ respectively), the lines $AL$ and $MB$ intersect inside the triangle.
We assumed in this case that the intersection point $P$ is at $\infty$, outside the triangle. By the above, at most one of the points $L,M,N$ is on the triangle. By our sign argument, it can't be zero points on the triangle. The only remaining alternative is that exactly one of the points is on the triangle. Done.
This is not the only argument. We can also angle-chase it with the parallel lines, arguing that exactly one of the parallel lines $AL$, $BM$, $CN$ lies inside the angle of the triangle at that vertex.
