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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 point $O$ when their three corresponding-vertex projective lines $P{P}^{\prime }$, $Q{Q}^{\prime }$, and $R{R}^{\prime }$ all pass through $O$.

Desargues's theorem: perspectivity from a point implies perspectivity from a line. More explicitly, if $P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }$ are concurrent, then the three corresponding-side intersectionsare collinear. The usual statement assumes nondegenerate projective triangles and well-defined, distinct corresponding-side intersections. The projective line containing $X,Y,Z$ is the axis of perspectivity from a line. All intersections are projective: parallel lines in an affine chart meet on its projective line at infinity.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

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.