Desargues's theorem ($P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }\text{ concurrent}⟹PQ\cap P^{\prime }{Q}^{\prime },QR\cap Q^{\prime }{R}^{\prime },RP\cap R^{\prime }{P}^{\prime }\text{ collinear}$)

For two projective triangles in a projective plane over a field, perspectivity from a point implies perspectivity from a line. The usual nondegenerate form assumes that the corresponding-side intersections are well defined and distinct. Its dual is the converse, by point-line duality. The result holds in the real projective plane; it is not an axiom valid in every abstract incidence plane.

One linear algebra proof explains the collinearity. Choose a representative $o$ of the center. Rescale the paired representatives to $p^{\prime }=p+\alpha o$, $q^{\prime }=q+\beta o$, and $r^{\prime }=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p^{\prime }-\alpha q^{\prime }$. Similarly choose $y=\gamma q-\beta r$ and $z=\alpha r-\gamma p$. Then $\gamma x+\alpha y+\beta z=0$, so the three representatives are linearly dependent. Their projective points are consequently collinear.

Two projective triangles are in perspectivity from a line when the three intersections of corresponding sides lie on one projective line, called their axis of perspectivity. In the nondegenerate setting this is dual to perspectivity from a point under point-line duality.

Two projective triangles are in perspectivity from a point when the three projective lines joining corresponding vertices meet in one projective point, called their center of perspectivity.