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,   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.

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