Proof: By Euclid

∎

For let the random points $F$ and $G$ have been taken on $EC$ and $EB$ (respectively).
And let $CB$ and $FG$ have been joined, and let $FH$ and $GK$ have been drawn across.
I say, first of all, that triangle $ECB$ is in one (reference) plane.
For if part of triangle $ECB$, either $FHC$ or $GBK$, is in the reference [plane], and the remainder in a different (plane) then a part of one the straight lines $EC$ and $EB$ will also be in the reference plane, and (a part) in a different (plane).
And if the part $FCBG$ of triangle $ECB$ is in the reference plane, and the remainder in a different (plane) then parts of both of the straight lines $EC$ and $EB$ will also be in the reference plane, and (parts) in a different (plane).
The very thing was shown to be absurb [Prop. 11.1].
Thus, triangle $ECB$ is in one plane.
And in whichever (plane) triangle $ECB$ is (found), in that (plane) $EC$ and $EB$ (will) each also (be found).
And in whichever (plane) $EC$ and $EB$ (are) each (found), in that (plane) $AB$ and $CD$ (will) also (be found) [Prop. 11.1].
Thus, the straight lines $AB$ and $CD$ are in one plane, and every triangle (formed using segments of both lines) is in one plane.
(Which is) the very thing it was required to show.

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The proofs of the first three propositions in this book are not at all rigorous. Hence, these three propositions should properly be regarded as additional axioms (translator's note) ↩