Proving a theorem using Pappus' theorem

Question

I need some help. I want to prove Desargues' theorem via using Pappus' theorem. And I don't know how. Please, help me!

Answer 1

The original work here would be Beweis des Desarguesschen Satzes aus dem Pascalschen by Hessenberg (1905), which is the reason why this is called Hessenberg's theorem.

My lecture notes suggest the following:

Given $AB\Vert A'B'$ and $BC\Vert B'C'$ you want to show $AC\Vert A'C'$. This is a Euclidean formulation of Desargues' theorem, and I'll also use Euclidean versions of Pappos' theorem to show it. You might of course replace all occurrences of the line at infinity by any other line, since all of this is invariant under projective transformations.

1. Draw $OP\Vert BC$ and intersect it with $AC$ to obtain $P$. Draw $PQ\Vert AB$ and intersect with $OB$ to obtain $Q$. Also draw $QC$. By Pappos' theorem on $OQB,PAC$ this is parallel to $OA$.
2. Construct $R$ as the intersection of $PA'$ with $B'C'$. The line $QC$ with pass through $R$ as well, due to Pappos' theorem on the points $OQB',PA'R$.
3. Now you have Pappos' theorem a third time on the points $OCC',PA'R$ which shows $AC\Vert A'C'$ as required.

Answer 2

Let me follow MvG's answer to prove the general case:

$\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ share the perspective center $O$ then we need to prove $G=A_1B_1\cap A_2B_2$, $H=B_1C_1\cap B_2C_2$ and $J=C_1A_1\cap C_2A_2$ are collinear.

1. Draw $P=A_1C_1\cap OH$, $Q=OB_1\cap GP$ and $K=OA_1\cap C_1Q$, then we know $GHK$ are collinear by Pappus on $OQB_1$ and $PA_1C_1$;
2. Draw $R=HB_2\cap KQ$, then we know $PA_2R$ are collinear by Pappus on $OQB_2$ and $GHK$;
3. $HJK$ are collinear by Pappus on $PA_2R$ and $OC_1C_2$;
4. $GHJ$ are collinear because of $GHK$ and $HJK$.

Because of the duality, this process also proves the converse of Desargues's theorem: just represent $A_1B_1C_1$ and $A_2B_2C_2$ as edges of the two triangles and $O$ as the perspective axis, and apply Pappus's dual in step 1~3.