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I have a question regarding the proof of Desargues' Theorem. When the traingles $ABC$ and $A'B'C'$ are assumed to be lying on the same plane.

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A point $X$ is taken outside that plane, and the lines $XA,XB,XC,XA',XB'$ and $XC'$ are drawn. Then $D$ is taken on the line $XB$, and the intersection of $OD$ and $XB'$ is taken to be $D'$.

Now the book says that $AB\cap A'B'=AD\cap A'D'$. I don't understand how that is.

For reference, please refer to pg. 8 (end of the page), Case 2 of this link

1 Answers1

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The way I read it, the book does not say that $AB\cap A'B'=AD\cap A'D'$. What it does say is this:

But these points are projected for $X$ into $P,Q,R$, on $\Sigma$, hence $P,Q,R$ are collinear.

Do you have some point $P'=AD\cap A'D'$ and $X,P,P'$ are collinear, so if you project $P'$ into $\Sigma$ with $X$ as the center of projection, then you end up at $P$. Likewise for $R'$ to $R$. So $P',Q,R',X$ lie in one plane, and that plane intersects $\Sigma$ in the line $P,Q,R$.

MvG
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  • Why do $P',Q,R'$ and $X$ need to lie on the same plane? –  Jul 04 '14 at 08:26
  • @Ayush: Because $P',Q,R'$ are collinear as per case 1, and because a line and a point span a plane. – MvG Jul 04 '14 at 08:27
  • @MvG- Just one more question. Why are $X,P,P'$ collinear? I thought this was because of case 1 too, but I'm not sure. –  Jul 04 '14 at 08:42
  • @Ayush: $X,P,P'$ are collinear because they lie in both the plane $XAB$ and $XA'B'$. And that in turn is because $D$ lies in plane $XAB$ because it lies on line $XB$, while $D'$ lies in $XA'B'$ because it lies on $XB'$. – MvG Jul 04 '14 at 09:11
  • I finally did understand. Thanks a lot for some much needed patient explaining. :) –  Jul 04 '14 at 12:05