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

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

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 $O$.

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$\blacksquare$

Also see

• Desargues' Theorem

Source of Name

This entry was named for Girard Desargues.

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Desargues' theorem
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Desargues' theorem
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Desargues' theorem