A doubt in the proof of Desargues' Theorem.

Question

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.

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

Answer 1

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 $P,Q,R$.