Simple projective geometry (in the projective plane) question

Question

Let, in the projective plane, $A,B,C,D,E$ and $F$ be the intersection points of the sides of a quadrilateral (4 lines, no 3 of which are concurrent with the same point) so that the sides are $ABF$, $ADE$, $DCF$ and $BCE$ and similarly $A',B',C',D',E'$ and $F'$ be the intersection points of the sides of a second quadrilateral so that the sides are $A'B'F'$, $A'D'E'$, $D'C'F'$ and $B'C'E'$. Let $\pi_1=ADE\barwedge A'D'E'$ be the unique projectivity mapping $A$, $D$ and $E$ onto $A'$, $D'$ and $E'$ respectively and similarly $\pi_2=DCF\barwedge D'C'F'$, $\pi_3=ABF\barwedge A'B'F'$ and $\pi_4=BCE \barwedge B'C'E'$. Suppose $l$ is a line not through $D$ and $B$, then $l$ meets $ADE$ in, say, $X_l$ and meets $DCF$ in, say, $Y_l$, they are different points and so are $\pi_1(X_l)$ and $\pi_2(Y_l)$. The same can be said for the intersection points of $l$ with $ABF$, say, $W_l$ and with $BCE$, say, $Z_l$ and their images $\pi_3(W_l)$ and $\pi_4(Z_l)$. Now, the fact that there exists a (necessarily unique) projective collineation mapping the first quadrilateral onto the second (with A to A' etc..) proves that the lines $\pi_1(X_l)\pi_2(Y_l)$ and $\pi_3(W_l)\pi_4(Z_l)$ are the same (in fact the image of $l$ under the projective collineation). I'm pretty sure that there exists a very simple way of proving this, but I'm not seeing it. Anyone with an idea? Thanks in advance.

Answer 1

Let $W$ be a point on $ABF$ and $\sigma_{W}$ be the perspectivity with centre $W$ that maps the points of $ADE$ onto the points of $DCF$. $\pi_2 \circ \sigma_{W} \circ \pi^{-1}_1$ is a projectivity that maps the points of $A'D'E'$ onto the points of $D'C'F'$ with invariant $D'$ hence it's a perspectivity $\sigma_{W'}$ with centre $W'$ (not on $A'D'E'$ and not on $D'C'F'$). $\sigma_{W'}(A')=F'$ hence the centre $W'$ lies on $A'B'F'$ and is not equal to $A',F'$. Note that, whenever $X$ is a point on $ADE$ and $Y$ a point on $CDF$ so that $X,Y$ and $W$ are collinear, then we have $\pi_2 \circ \sigma_{W} \circ \pi^{-1}_1(\pi_1(X))=\pi_2(Y)$ hence $W'$ lies on $\pi_1(X)\pi_2(Y)$, so $W'$ is the intersection of $\pi_1(X)\pi_2(Y)$ and $A'B'F'$.

Set $\eta(W)=W'$. The above definition does not work for $A$ and $F$, we set $\eta(A)=A'$ and $\eta(F)=F'$. Let $X$ be a random point on $ADE$, $X \neq A,D$ and let $\sigma_X$ be the perspectivity with centre $X$ mapping points of $ABF$ to the points on $DCF$ and $\sigma_{\pi_1(X)}$ be the perspectivity with centre $\pi_1(X)$ mapping the points on $D'C'F'$ to the points on $A'B'F'$. $\sigma_{\pi_1(X)}\circ \pi_2 \circ \sigma_{X}$ maps a point $W\neq A,F$ on $ABF$ onto the intersection of $\pi_1(X)\pi_2(\sigma_{X}(W))$ with $A'B'F'$. Since $W, X$ and $\sigma_{X}(W)$ are collinear we have $\eta(W)=\sigma_{\pi_1(X)}\circ \pi_2 \circ \sigma_{X}(W)$. Also $\eta(A)=\sigma_{\pi_1(X)}\circ \pi_2 \circ \sigma_{X}(A)=A'$ and $\eta(F)=\sigma_{\pi_1(X)}\circ \pi_2 \circ \sigma_{X}(F)=F'$. Hence $\eta$ is the projectivity $\sigma_{\pi_1(X)}\circ \pi_2 \circ \sigma_{X}$ with $X$ a random point on $ADE$ not equal to $A,D$. Now take $E$ for $X$. We have $\sigma_{\pi_1(E)}\circ \pi_2 \circ \sigma_{E}(A)=A'$, $\sigma_{\pi_1(E)}\circ \pi_2 \circ \sigma_{E}(B)=B'$ and $\sigma_{\pi_1(E)}\circ \pi_2 \circ \sigma_{E}(F)=F'$. Hence $\sigma_{\pi_1(E)}\circ \pi_2 \circ \sigma_{E}=\pi_3$ so $\eta = \pi_3$.

We now have everything we need: $X_l$ on $ADE$, $Y_l$ on $DCF$ and $W_l$ on $ABF$ collinear, so $\eta(W_l)=\pi_3(W_l)$ on $\pi_1(X_l)\pi_2(Y_l)$. We can prove that $\pi_4(Z_l)$, $\pi_1(X_l)$ and $\pi_2(Y_l)$ are collinear in the same way.