Straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other.

Let each of the straight lines *AB* and *CD* be parallel to *EF,* but not in the same plane with it.

I say that *AB* is parallel to *CD.*

Let a point *G* be taken at random on *EF,* and from it draw *GH* in the plane through *EF* and *AB* at right angles to *EF,* and *GK* in the plane through *EF* and *CD* again at right angles to *EF.*

Now, since *EF* is at right angles to each of the straight lines *GH* and *GK,* therefore *EF* is also at right angles to the plane through *GH* and *GK.*

And *EF* is parallel to *AB,* therefore *AB* is also at right angles to the plane through *HG* and *GK.*

For the same reason *CD* is also at right angles to the plane through *HG* and *GK.* Therefore each of the straight lines *AB* and *CD* is at right angles to the plane through *HG* and *GK.*

But if two straight lines are at right angles to the same plane, then the straight lines are parallel. Therefore *AB* is parallel to *CD.*

Therefore, *straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other.*

Q.E.D.

Consider a quadrilateral ABCD whose four vertices may or may not lie in a plane. Let E, F, G, and H be the midpoints of the sides AB, BC, CD, and DA, respectively. Then the quadrilateral EFGH lies in a plane and is a parallelogram, called the Varignon parallelogram. Varignon (1654–1722) showed the area of a planar quadrilateral is twice the area of this parallelogram.
The proof that As a corollary, it follows that the lines joining the midpoints of an arbitrary quadrilateral are concurrent and bisect each other, even if the four sides of the quadrilateral do not lie in a plane. (These are the lines |