Proving converse of Desargues Consider $\triangle ABC$ and $\triangle A'B'C'$. Set $E=AB\cap A'B'$, $F=BC\cap B'C'$, and $G=AC\cap A'C'$. I want to show that if $E$, $F$, and $G$ are collinear, then $\triangle ABC$ and $\triangle A'B'C'$ are in perspective from some point $O$.
I realize that the converse of Desargues is the dual of the original theorem and can be proven easily in this way.
However, I am wondering how to prove the converse a bit more directly (but not the most direct way).
Specifically, can we view the line $EFG$ as the line at infinity in $\mathbb {RP}^2$? I have heard about such a technique being useful. But how does it work?
 A: If you accept the idea that for any two lines in the projective plane, theres a projective map sending one to the other, then as a first step, you can find a map $f$ that takes $E, F, G$ to the line at infinity. Letting 
$a = f(A), b = f(B), $ etc., you then prove the theorem is true for $a, b, c, a', b', c'$, etc., resulting in some point $o$, and then conclude that it must be true by letting $O = f^{-1}(o)$ and showing that $ABC$ and $A'B'C'$ are perspective from $O$ (which is straightforward). 
Now what about the proof for the lowercase letters?
[Oops...I just realized that in everything I've written below, I should have used lowercase rather than uppercase...I trust you can translate.]
Well, in this case knowing that $AB$ and $A'B'$ meet at $E$, which is a point at infinity, you know that $AB$ and $A'B'$ are parallel, and similarly for the other two sides. So $ABC$ and $A'B'C'$ are similar triangles, with parallel corresponding sides. The degenerate case, in which $C$ is on $AB$, isn't very interesting, and I'll let you work out the details for yourself. So I'm assuming henceforth that the triangles are nondegenerate. 
Case 1: They're identical. In this case, they're perspective from any point $O$ and we're done. 
Case 2: $A = A'$, but $B \ne B'$ and $C \ne C'$ (or any permutation of this). Then the triangles are perspective from $A$. 
Case 2A: $A = A'$, $B = B'$, but $C \ne C'$. This case can't happen, but I'll let you figure out why. 
Case 3: $A \ne A', B \ne B', C \ne C'$, and $A'B'C'$ is just a translated copy of $ABC$. In this case, consider the line $\ell$ from $A$ to $A'$; this meets the line at infinity in a point $O$, and $A'B'C'$ is perspective to $ABC$ from $O$. 
Case 4: Same as 3, but the length of $A'B'$ is different from the length of $AB$. Then there's a constant $s$ and a vector $v$ with the following properties: letting $P$ (resp. $P'$) denote the center of $ABC$, we have $v = P' - P$, s = $\|A'-B'\|/\|A-B\|$ $A' = P' + s(A-P)$, $B' = P' + s(B-P)$, $C' = P' + s(C-P)$. (Why? Translate $A'B'C'$ by $-v$ so that the centers of $ABC$ and $A'B'C'$ match. Then by scaling around those (matching) centers by a scale factor $s$, we make $ABC$ match $A'B'C'$). 
At this point, working in coordinates, you can actually show that $A'A$, $B'B$, and $C'C$ intersect at some point. I'm pretty sure you can work out the location of that point ... it's something like $P - \frac{1}{s-1}v$, but I'm just spitballing here and have no idea whether I've gotten that right. 
And then you're done. 
Notice that in the case $s = 1$, you're really back to case $2$, and the divide-by-zero gets you the point at infinity that you get by intersecting $\ell$ with the line at infinity, etc. In other words, by separating into cases like this, and treating the line at infinity as special, we're actually making things harder by introducing more cases. Still, it does the job. 
