Desargues' Theorem/Converse

Theorem
Let $\triangle ABC$ and $\triangle A'B'C'$ be triangles lying in the same or different planes.

Let:
 * $BC$ meet $B'C'$ in $L$
 * $CA$ meet $C'A'$ in $M$
 * $AB$ meet $A'B'$ in $N$

where $L, M, N$ are collinear.

Then the lines $AA'$, $BB'$ and $CC'$ intersect in the point $O$.

Proof

 * DesarguesTheorem.png

Let $L$, $M$ and $N$ be collinear.

Then $\triangle BB'N$ and $\triangle CC'M$ are perspective with center $L$ ($L = BC \cap B'C' \cap MN$)

From Desargues' Theorem:
 * $O = BB' \cap CC'$
 * $A = BN \cap CM$
 * $A' = C'M \cap B'N$

are collinear.

Thus:
 * $AA' \cap BB' \cap CC' = O$

Hence $\triangle ABC$ and $\triangle A'B'C'$ are perspective with center $O$.

Also see

 * Desargues' Theorem