Elementary proof of a triangular grid lemma I am looking for an elementary proof of the following lemma, which concerns what Green and Tao call "triangular grids" (see arXiv:1208.4714).
Let $a_1$, $a_2$, $a_3$, $a_4$, $b_1$, $b_2$ be six arbitrary lines in the plane (in general position, for whatever the appropriate sense of "general position" is).
Let $c_1$ be the line through the intersection points $a_3b_1$ and $a_2b_2$.
Let $c_2$ be the line through $a_4b_1$ and $a_3b_2$.
Let $b_3$ be the line through $a_1c_1$ and $a_2c_2$.
Let $c_3$ be the line through $a_4b_2$ and $a_3b_3$.
Let $b_4$ be the line through $a_1c_2$ and $a_2c_3$.
Let $c_4$ be the line through $a_4b_3$ and $a_3b_4$.
Let $b_5$ be the line through $a_1c_3$ and $a_2c_4$.
Let $c_5$ be the line through $a_4b_4$ and $a_3b_5$.
Then the intersection points $b_1c_3$, $b_2c_4$, $b_3c_5$ are collinear.
As a consequence, this triangular grid can be indefinitely continued.
See figure below.
This can be proven using cubic curves, specifically the Cayley-Bacharach theorem: In the dual setting, all the points dual to the given lines lie on a common cubic curve. See the above-mentioned paper of Green and Tao.
But as I said, I was wondering whether an elementary proof exists.
I see there is previous literature on these triangular grids (also called with other names; see reference [6] in the above paper, and references cited there). I quickly glanced at the literature, and I didn't find what I'm looking for.

 A: We can prove this with a cross ratio chase, but first we need an easy lemma.

Lemma: If points $A,B,C,D$ are on one line and $E,F,G,H$ are on another line, then $(A,B;C,D) = (E,F;G,H)$ if and only if the three points $X = AF\cap BE, Y = BG\cap CF, Z = CH\cap DG$ lie on a line.
Proof: Let $P = AG\cap CE, Q = CG\cap XY$. By Pappus's Theorem, $P$ is on line $XY$. Projecting through $G$, we have $(A,B;C,D) = (P,Y;Q,DG\cap XY)$, and projecting through $C$, we have $(E,F;G,H) = (P,Y;Q,CH\cap XY)$. Thus $(A,B;C,D) = (E,F;G,H)$ if and only if $CH, DG$, and $XY$ meet at a point.

Now we need to name some points. Let $A = a_2\cap b_1, B = a_3\cap b_1\cap c_1, C = a_4\cap b_1\cap c_2, D = b_1\cap c_3, E = b_1\cap c_4$, let $F = a_1\cap b_3\cap c_1, G = a_2\cap b_3\cap c_2, H = a_3\cap b_3\cap c_3, I = a_4\cap b_3\cap c_4, J = b_3\cap c_5$, and let $K = b_5\cap c_2, L = a_1\cap b_5\cap c_3, M = a_2\cap b_5\cap c_4, N = a_3\cap b_5\cap c_5, O = a_4\cap b_5$. Finally, let $X = AN \cap DK$.
By two visually obvious applications of the Lemma we have $(A,B;C,D) = (F,G;H,I) = (K,L;M,N)$. Thus $(B,D;A,C) = (L,N;K,M)$, so by a slightly less obvious application of the Lemma $X$ must be on the line $GH = b_3$. Applying the Lemma again, since $X$ is on the line $GI = b_3$ we have $(E,C;A,D) = (O,M;K,N)$.
Putting $(A,B;C,D) = (K,L;M,N)$ and $(E,C;A,D) = (O,M;K,N)$ together, we see that $(B,C;D,E) = (L,M;N,O)$ (consider the projective transformation from line $b_1$ to line $b_5$ taking $A,C,D$ to $K,M,N$: the first equality says it takes $B$ to $L$, and the second says it takes $E$ to $O$). Applying the Lemma in another visually obvious way, we have $(L,M;N,O) = (G,H;I,J)$. Thus $(B,C;D,E) = (G,H;I,J)$, so by the Lemma the intersection $DJ\cap EI$ is on the line connecting $BH\cap CG, CI\cap DH$, which is $b_2$. Since $D = b_1\cap c_3$, $J = b_3\cap c_5$, $EI = c_4$, this means that $b_1\cap c_3, b_2\cap c_4, b_3\cap c_5$ are collinear.
