Prove that 3 lines are concurrent

Question

I've been stuck with the following problem that I found in a projective geometry book, which I'm self-studying.

Assume we have in $\mathbb{P}_{2}$ a triangle with vertices $A,B,C$ and two different points $M, N$, no one lying on a side of the triangle. Let $A', B', C'$ be, respectively, the intersections of the line $MN$ with the sides of the triangle opposite $A, B, C$. Denote by $\overline{A}, \overline{B}, \overline{C}$ the fourth harmonics of $A', B'$ and $C'$ with respect to $M, N$.

Prove that the lines $A\overline{A}, B\overline{B}, C\overline{C}$ are concurrent.

I've tried using a reference to find the different fourth harmonic points and from there derive the equations of the lines, but nothing seems to work out. Any help would be highly appreciated. Thanks.

Answer 1

Hint

Since $A,B,C$ are collinear thanks to (unsigned) Menelaus' Theorem you can write $$ \frac{|AB'|}{|B'A|} + \frac{|CA'|}{|A'C|} + \frac{|BC'|}{|C'B|} = 1 $$ Then using the definition of Harmonic points you have $$ \frac{|AB''|}{|B''A|} + \frac{|CA''|}{|A''C|} + \frac{|BC''|}{|C''B|} = 1 $$ Hence Ceva Theorem gives you the concurrence.

Answer 2

Another proof.

Let line $l$ be the line determined by $MN$. Let the intersection point of line $B\overline{B}$ with line $CA$ be $K$.

Now consider the two projections $p_A$ and $p_C$ (are they called perspectivities or projecitivties?) of line $l$ onto line $B\overline{B}$ from points $A$ and $C$ respectively. Let $A\overline{A}$ intersect $B\overline{B}$ at point $P_A$ and let $C\overline{C}$ intersect $B\overline{B}$ at point $P_C$. Then in terms of projections, $P_A = p_A(\overline{A})$ and $P_C = p_C(\overline{C})$. Since the projective transformations $p_A$ and $p_C$ preserve cross-ratios, the cross ratio of points $\overline{B}, B', \overline{A}, C'$ equals the cross-ratio of their images with respect to $p_A$ which are $\overline{B}, K, P_A, B$, and furthermore the cross ratio of $\overline{B}, B', \overline{C}, A'$ equals the cross-ratio of their images with respect to $p_C$ which are $\overline{B}, K, P_C, B$. However, due to the fact that the points $\overline{A}, \overline{B}, \overline{C}$ are the harmonic images of the corresponding points $A', B', C'$ with respect to the pair of points $M, N$, one concludes that the cross-ratios of $\overline{B}, B', \overline{A}, C'$ and $\overline{B}, B', \overline{C}, A'$ are equal. Consequently, the cross-ratios $\overline{B}, K, P_A, B$ and $\overline{B}, K, P_C, B$ are also equal which is possible if and only if $P_A \equiv P_C \equiv P$. Therefore, the three lines $A\overline{A}, B\overline{B}, C\overline{C}$ intersect at the common point $P$.