**Theorem
2.9**: Let * A*, *B*, and *C *be three noncolinear points, let *D* be a point on the line segment strictly between *A* and *B*, and let *E* be a point on the line segment strictly between
*A* and *C*. Then *DE* is parallel to *BC* if and
only if there is a nonzero real number * t* such that

and

**Proof**: The requirement that * t* be nonzero follows from the
fact that both * D* and *E* are distinct from *A*.

Assume that there is such a real number *t*. Let

Then if we assume that there is a nonzero real number *t* such that

and

then

and

We must first dispose of the case where * BC* is vertical. In that case
*x*_{1} = *x*_{2}, but then, *D* and *E* can be seen to have
the same *x*-coordinates as well, so *DE* will also be vertical, and thus be
parallel to *BC*.

Otherwise, the slope of
the line between * D* and *E* is

When we remove the parentheses in the top and bottom, the expression will simplify to

We can now factor

Since *t* is nonzero, we can cancel and get

which is the same as the
slope of the
line between *B* and *C*. As a
result, by the definition of parallel
lines, we conclude that the
line between * D* and *E* is parallel to the
line between *B* and *C*.

The converse follows from the uniqueness of the
line through *D* parallel to *BC, *Theorem 1.7. We know
that the line through * D* which contains the point * E* is parallel to *BC*. By Theorem 1.7. any
line through *D* parallel to *BC* must
then be the line
determined by *DE*.