Desargues' Theorem with Proof

Theorem 

Let $△ABC$ and $△A′B′C′$ 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 $M$, $AB$ meets $A′B′$ in $N$ where $L,M,N$ are collinear. 

Converse 

Let $△ABC$ and $△A′B′C′$ 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 $△ABC$ and $△A′B′C′$ be in different planes $π$ and $π′$ 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 $π$ and $π′$.

Now let $△ABC$  and $△A′B′C′$  be in the same plane $π$. Let $OPP′$  be any line through $O$  which does not lie in $π$ . 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 $B′′C′′$  are the three lines of intersection of the three planes $PBC , P′B′C′$  and π taken in pairs. So $BC , B′C′$  and $B′′C′′$ meet in a point $L$. 

Similarly $CA , C′A′$  and $C′′A′′$  meet in a point $M$  and $AB , A′B′$  and $A′′B′′$ meet in a point $N$. The two triangles $△ABC$  and $△A′′B′′C′′$  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. 

From Proofwiki