= Desargues's theorem
{c}
{title2=$PP',QQ',RR'\text{ concurrent}\Longrightarrow PQ\cap P'Q',QR\cap Q'R',RP\cap R'P'\text{ collinear}$}
= Desargues' theorem
{c}
{synonym}
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 $p'=p+\alpha o$, $q'=q+\beta o$, and $r'=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p'-\alpha q'$. 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>.
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