Let the two straight lines AB and BC meeting one another be parallel to the two straight lines DE and EF meeting one another not in the same plane.
I say that the plane produced through AB and BC and the plane produced through DE and EF do not meet one another.
Draw BG from the point B perpendicular to the plane through DE and EF to where it meets the plane at the point G.
Draw GH through G parallel to ED, and GK parallel to EF.
Now, since BG is at right angles to the plane through DE and EF, therefore it makes right angles with all the straight lines which meet it and lie in the plane through DE and EF.
But each of the straight lines GH and GK meets it and lies in the plane through DE and EF, therefore each of the angles BGH and BGK is right.
And, since BA is parallel to GH, therefore the sum of the angles GBA and BGH equals two right angles.
But the angle BGH is right, therefore the angle GBA is also right. Therefore GB is at right angles to BA. For the same reason GB is also at right angles to BC.
Since then the straight line GB is set up at right angles to the two straight lines BA and BC which cut one another, therefore GB is also at right angles to the plane through BA and BC.
But planes to which the same straight line is at right angles are parallel, therefore the plane through AB and BC is parallel to the plane through DE and EF.
Therefore, if two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel.