Prove that 3 points are collinear - Geometry problem

Question

If P, Q , R are interior points of the sides BC, CA and AB of a triangle ABC and the corresponding Cevians AP, BQ and CR are concurrent at a point X .

We have :

$\frac{AX}{XP}=\frac{AR}{RB}+\frac{AQ}{QC}$ ( Van Aubel's theorem )

We suppose that :

(RQ) and (BC) are concurrent at a point L

(PR) and (CA) are concurrent at a point M

(PQ) and (BA) are concurrent at a point N

How can I prove that L,M,N are collinear ?

Thanks in advance

Answer 1

This is a special case of the theorem of Desargues.

Desargue's theorem, wiki page

In our case, $X$ is the center of the perspective of the two triangles (with corresponding vertices on the same places in the writing)

• $\Delta ABC$,
• $\Delta PQR$.

(If this is not enough, i may try to give a proof adapted to the given situation, where we also have harmonic proportions on the side lines $BC$, $CA$, $AB$ by the given those points on each line.)

Answer 2

Comment:

As can be seen in figure point X and Line A'C' is common for triangles ABC and PQR, this may help.