Projective geometry

A projective transformation is represented by an invertible linear map on homogeneous coordinates, and the cross-ratio is built only from coefficients expressing two of the points as linear combinations of the other two; since a linear map preserves linear combinations exactly, it cannot change those coefficients, so the cross-ratio survives.

Represent the four collinear points by homogeneous coordinate vectors $A, B, C, D \in \mathbb{R}^3$ lying in a common 2-dimensional subspace, and represent $\varphi$ by an invertible $3\times 3$ matrix $M$, so that $\varphi$ applied to a point with vector $P$ has homogeneous coordinates $MP$.

Because $A, B, C, D$ are collinear, representatives can be chosen so that $C = \lambda_1 A + \mu_1 B$ and $D = \lambda_2 A + \mu_2 B$ for scalars $\lambda_1, \mu_1, \lambda_2, \mu_2$; the cross-ratio $(A,B;C,D)$ is defined purely in terms of these four scalars, specifically as $\dfrac{\mu_1/\lambda_1}{\mu_2/\lambda_2}$.

Applying the linear map $M$ gives $MC = \lambda_1(MA) + \mu_1(MB)$ and $MD = \lambda_2(MA) + \mu_2(MB)$, because matrix multiplication distributes over linear combinations; so the very same scalars $\lambda_1, \mu_1, \lambda_2, \mu_2$ express $\varphi(C)$ and $\varphi(D)$ in terms of $\varphi(A)$ and $\varphi(B)$.

Since the cross-ratio depends only on these scalars and they are unchanged by $M$, we conclude $(\varphi(A),\varphi(B);\varphi(C),\varphi(D)) = (A,B;C,D)$, which is exactly the claimed invariance.

Working purely inside the flat plane, the claim looks delicate because it mixes many different lines. Lifting the picture into three-dimensional space turns every incidence into an intersection of planes, and two distinct planes always meet in a line, which forces the three points onto one common line.

Regard the two triangles as lying in a plane $\pi$ inside $\mathbb{R}^3$, and construct, above $\pi$, a spatial configuration: a point $\widehat{O}$ not in $\pi$, and two triangles $\widehat{A}\widehat{B}\widehat{C}$ and $\widehat{A'}\widehat{B'}\widehat{C'}$ in two different planes through space, positioned so that lines $\widehat{A}\widehat{A'}$, $\widehat{B}\widehat{B'}$, $\widehat{C}\widehat{C'}$ all pass through $\widehat{O}$, and so that projecting this spatial picture back onto $\pi$ recovers the original triangles $ABC$ and $A'B'C'$ in perspective from $O$.

In this spatial picture, the line $\widehat{B}\widehat{C}$ and the line $\widehat{B'}\widehat{C'}$ both lie in the plane spanned by $\widehat{O}, \widehat{B}, \widehat{C}$ (since $\widehat{B'}$ and $\widehat{C'}$ lie on lines through $\widehat{O}$ and $\widehat{B}, \widehat{C}$ respectively), so these two lines meet at a point $\widehat{P}$; similarly $\widehat{Q}$ and $\widehat{R}$ exist in space as the analogous intersections for sides $CA$ and $AB$.

The three points $\widehat{P}, \widehat{Q}, \widehat{R}$ all lie in the plane containing triangle $\widehat{A}\widehat{B}\widehat{C}$, since each is built from a pair of its sides, and simultaneously in the plane containing $\widehat{A'}\widehat{B'}\widehat{C'}$. Hence $\widehat{P}, \widehat{Q}, \widehat{R}$ lie on the line where these two planes intersect, because any two distinct planes in space meet in exactly one line.

Projecting this spatial configuration back down onto the plane $\pi$ sends $\widehat{P}, \widehat{Q}, \widehat{R}$ to $P, Q, R$ and preserves collinearity, since a projection from a point sends any line to a line. Therefore $P, Q, R$ are collinear in $\pi$, which is exactly Desargues' theorem.