Desargues' Theorem

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

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

Then $BC$ meets $B'C'$ in $L$, $CA$ meets $C'A'$ in $M$ and $AB$ meets $A'B'$ in $N$, where $L, M, N$ are collinear.

Proof
Let $\triangle ABC$ and $\triangle A'B'C'$ be in different planes $\pi$ and $\pi'$ respectively.

Since $BB'$ and $CC'$ intersect in $O$, it follows that $B$, $B'$, $C$ and $C'$ lie in a plane.

Thus $BC$ must meet $B'C'$ in a point $L$.

By the same argument, $CA$ meets $C'A'$ in a point $M$ and $AB$ meets $A'B'$ in a point $N$.

These points $L, M, N$ are in each of the planes $\pi$ and $\pi'$.

By Two Planes have Line in Common they are therefore collinear on the line where $\pi$ and $\pi'$ meet.

Now let $\triangle ABC$ and $\triangle A'B'C'$ be in the same plane $\pi$.

Let $OPP'$ be any line through $O$ which does not lie in $\pi$.

Then since $PP'$ meets $AA'$ in $O$, the four points $P, P', A, A$ are coplanar.

Thus $PA$ meets $P'A'$ at a point $A''$.

Similarly $PB$ meets $P'B'$ at a point $B$, and $PC$ meets $P'C'$ at a point $C$.

The lines $BC, B'C'$ and $BC$ are the three lines of intersection of the three planes $PBC$, $P'B'C'$ and $\pi$ taken in pairs.

So $BC$, $B'C'$ and $BC$ meet in a point $L$.

Similarly $CA$, $C'A'$ and $CA$ meet in a point $M$ and $AB$, $A'B'$ and $AB$ meet in a point $N$.

The two triangles $\triangle ABC$ and $\triangle ABC$ are in different planes, and $AA$, $BB$ and $CC$ meet in $P$.

Thus $L$, $M$ and $N$ are collinear by the first part of this proof.