Desargues' Theorem

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.