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

D = (1 - t)A + tB

and

E = (1 - t)A + tC

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

A = (x0, y0)

B = (x1, y1)

C = (x2, y2)

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

D = (1 - t)A + tB

and

E = (1 - t)A + tC

then

D = ((1 - t)x0 + tx1, (1 - t)y0 + ty1)

and

E = ((1 - t)x0 + tx2, (1 - t)y0 + ty2)

We must first dispose of the case where BC is vertical. In that case x1 = x2, 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. 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.

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