Theorem 2.10: Let A, B, and C be three noncolinear points. If D is on the line through A which is parallel to BC then there is a real number s such that

D = A + s(C - B)

Proof: If A = D, then we can take s = 0, and we're done. So we can assume that D is a different point than A.

Let

A = (x0, y0)

B = (x1, y1)

C = (x2, y2)

and

D = (x, y)

Here we are considering the case where the line is neither horizontal or vertical. Since the lines are neither horizontal nor vertical, saying that D is on a line through A parallel to BC is equivalent to

Multiply both sides by x - x0 and divide both sides by y2 - y1.

Since the line is neither horizontal nor vertical, both sides of this equation are defined.

Define

Then

A + s(C - B)

= A - sB + sC

= (x0, y0) - s(x1, y1) + s(x2, y2)

= (x0 - sx1 + sx2, y0 - sy1 + sy2)

= (x0 + s(x2 - x1), y0 + s(y2 - y1))

= (x0 + x - x0, y0 + y - y0))

= (x, y)

= D

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horizontal case

vertical case

next theorem (2.11)