Cross-ratio concept from Euclidean Geometry

Question

I have seen on the following Wikipedia website the definition of "cross-product." Apparently, if four lines passing through a common point $P$ are traversed by a line and the points of intersection "in one direction" are A, B, C, and D, their cross-product is \begin{equation*} [A, B; C, D] = \frac{\mathit{AC}\cdot \mathit{BD}}{\mathit{AD} \cdot \mathit{BC}} . \end{equation*} Why is this quantity defined this way? Is there supposed to be a semicolon in this notation? If another line intersects the same four lines passing through $P$ at $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, and $D^{\prime}$ "in the same direction," \begin{equation*} [A, B; C, D] = [A^{\prime}, B^{\prime}; C^{\prime}, D^{\prime}] . \end{equation*} What is a proper explanation for this? What happens if the points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, and $D^{\prime}$ "in the opposite direction" with respect to $A$, $B$, $C$, and $D$?

https://en.wikipedia.org/wiki/Cross-ratio

May someone offer me a textbook in Euclidean Geometry that has a discussion on cross-product?

Answer 1

The cross-ratio of four points $A,B,C$ and $D$ on a line $l$ in Euclidean space is defined as the ratio of the ratios $AC:BC$ and $AD:BD$. Simplifying the double quotient yields $$ [A,B;C,D] = \frac{\frac{AC}{BC}}{\frac{AD}{BD}} = \frac{AC\cdot BD}{AD \cdot BC}.$$ The semicolon in Wikipedia's notation indicates the different roles of $A,B$ and $C,D$ here.

Now, if $P$ is a fifth point with distance $d$ to $l$, we can express $[A,B;C,D]$ via the areas of the triangles $[PAC]$, $[PBC]$, $[PAD]$ and $[PBD]$: $$ [A,B;C,D] = \frac{\frac{AC \cdot d}{2} \cdot \frac{ BD \cdot d}{2}}{\frac{AD \cdot d }{2}\cdot \frac{BC \cdot d}{2}} = \frac{[PAC] \cdot [PBD]}{[PAD] \cdot [PBC]},$$ which by the sine area theorem can be written as $$ [A,B;C,D] = \frac{PA \cdot PC \cdot \sin(PA,PC) \cdot PB \cdot PD \cdot \sin(PB,PD)}{PA \cdot PD \cdot \sin(PA,PD) \cdot PB \cdot PC \cdot \sin(PB,PC)} = \frac{\sin(PA,PC) \cdot \sin(PB,PD)}{\sin(PA,PD) \cdot \sin(PB,PC)}.$$ where $\sin(PX,PY)$ denotes the oriented angle between the lines $PX$ and $PY$. Remarkably, this expression does only depend on the angle between the four lines $PA$, $PB$, $PC$ and $PD$, and not at all on how $l$ intersects these. Therefore, if $A', B', C'$ and $D'$ are four different points on $PA,PB,PC$ and $PD$ on another line $l'$, we will have $$[A,B;C,D] = \frac{\sin(PA,PC) \cdot \sin(PB,PD)}{\sin(PA,PD) \cdot \sin(PB,PC)} = \frac{\sin(PA',PC') \cdot \sin(PB',PD')}{\sin(PA',PD') \cdot \sin(PB',PC')} = [A',B';C',D'].$$ Note that if $A$ and $A'$ lie on different sides of $P$, both $\sin(PA,PC)$ and $\sin(PA,PD)$ change sign, thus leaving the crossratio invariant.

A very practically oriented chapter on Euclidean cross-ratios would be in Evan Chen's "Euclidean Geometry in Mathematical Olympiads". Of course, one can study projective geometry much more abstractly, which is detailed in most textbooks on linear algebra.