If two harmonic quartets have a common point, prove their lines are concurrent Let $A,B,C,D$ and $A,L,M,N$ be collinear points such that $\{AB,CD\} = \{AL,MN\} = -1$. Prove that the lines BL, CN and DM concur.
I tried to build a triangle using A as a common point and then use Ceva Theorem, but the concurrent lines aren't the same in the triangle and the ones I have to use. Any other idea I could use?

A perspectivity, i.e. a projection from one line onto another through a common center, preserves cross ratios.

Consider the point where $BL$ and $CM$ intersect. Since the lines intersect in $A$, this perspectivity will map $A$ to itself. So you have three points $A,B,C$ and their images $A,L,M$. Therefore the image of the fourth point, $D$, has to be such that the cross ratio is preserved, i.e. has to be $N$.
Note that I'm using $CM$ and $DN$, while your question asks about $CN$ and $DM$. For arbitrary cross ratios, this distinction is important, and the other concurrence would be incorrect. For harmonic sets, i.e. cross ratio $-1$, you can swap $C$ and $D$ without affecting the value of the cross ratio. So you need this in your case.

