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

Antonio Riveira
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  • It would be helpful to include a statement of these two theorems. – Semiclassical Jul 16 '14 at 19:41
  • Two triangles are in perspective axially if and only if they are in perspective centrally.

    Denote the three vertices of one triangle by a, b, and c, and those of the other by A, B, and C. Axial perspectivity means that lines ab and AB meet in a point, lines ac and AC meet in a second point, and lines bc and BC meet in a third point, and that these three points all lie on a common line called the axis of perspectivity. Central perspectivity means that the three lines Aa, Bb, and Cc are concurrent, at a point called the center of perspectivity Deasrgues'theorem

     – Antonio Riveira Jul 16 '14 at 19:45
  • Great. Put that into the question. – Semiclassical Jul 16 '14 at 19:45
  • given one set of collinear points A, B, C, and another set of collinear points a, b, c, then the intersection points X, Y, Z of line pairs Ab and aB, Ac and aC, Bc and bC are collinear, lying on the Pappus line. It's Pappus'theorem – Antonio Riveira Jul 16 '14 at 19:46
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    Try getting hold of Casse's book Projective Geometry: an Introduction and look at exercise 3.4(2): he gives a number of intermediate questions and hints to show the Desagues theorem from the Pappus theorem. – Gro-Tsen Jul 16 '14 at 19:59
  • Thank you, I proved it without the book! And it's so easy, i just can't belive how it didn't cross my mind earlier. But now I have another problem, which is harder. How to prove the opposite of Desargues' theorem... i.e if the triangles are in perspective centrally, thus they are in perspective axially. :) – Antonio Riveira Jul 17 '14 at 19:13

2 Answers2

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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:

Figure

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

enter image description here

$\triangle A_1B_1C_1$ 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.

auntyellow
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