Showing whether the lines meet at a common unique point or not

Question

As shown in the figure below, there are two non parallel lines with four points on each line. If given that $(A,B;C,D)= (A',B';C',D')$, then is it necessary that the lines joining $A$ to $A'$, $B$ to $B'$, $C$ to $C'$, and $D$ to $D'$ all intersect at a single common unique point? If yes, how can one prove it?

• The quantity $(A, B; C, D)$ refers to the cross-ratio, defined by $$(A, B; C, D) = \frac{AC \cdot BD}{AD \cdot BC},$$ where $AC$, for example, is the distance between $A$ and $C$.
• Note: I thought of this when trying to see if the converse is true of theorem of cross-ratios (or not).

Answer 1

If the lines joining corresponding points intersect at a common point, then there is a bijective mapping between the lines $AD$ and $A’D’$ called a perspectivity. Perspectivities preserve cross ratios, but the converse is not true.

From the article:

The composition of two perspectivities is, in general, not a perspectivity. A perspectivity or a composition of two or more perspectivities is called a projectivity (projective transformation, projective collineation and homography are synonyms).

In the case of your diagram, the connecting lines will likely not be concurrent, but given that cross ratios are preserved you can come up with two perspectivities whose composition is your mapping.

Here's an example where the cross ratios are both equal to $2$.