Theorem 2.8: If a line segment contains points on both sides of another line, then the line must intersect the segment somewhere between its endpoints.
Proof: Let the two points be A = (x1, y1) and B = (x2, y2), and let the line be given by the equation ax + by = c.
If A is on one side and B is on the other, say
ax1 + by1 < c and ax2 + by2 > c
r = c - (ax1 + by1)ax2 + by2 and s = (ax2 + by2) - c
Note, both r and s are positive.
Note, 0 < t < 1, and
C = (1 - t)A + tB
is on the line segment between A and B by Theorem 2.1
C = (1 - t)(x1, y1) + t(x2, y2)
Let us substitute the coordinate of C into the equation of the line.
and we see that point C is on the line.
If the line determined by the segment doesn't cross the line then by Theorem 1.6, they are parallel. But if they are parallel, then by Theorem 2.6, all of the points of the segment are on the same side of the line, which contradicts the hypotheses. Since the lines are not parallel, then by Theorem 1.6, there will be a point of intersection between the two lines.
Suppose the point of intersection is not between the endpoints of the segment. Let A and B be the endpoints of the segment, and let C be the point of intersection. We can assume that A and B have been labeled so that C is on the opposite side of A from B.
for some negative real number t by Theorem 2.1 and the definition of the opposite side of A from B.
Since t is negative, 1 - t is greater than 1. Since it is not 0, we can divide both sides by it and get
If we let
So A is of the form
and since 1 - t is greater than 1, u is between 0 and 1, so A is on the segment between B and C.
In particular, A and B are on the same side of C, so by Theorem 2.5 they are on the same side of the line. By Theorem 2.7 any point on the line segment between them will be on the same side of the line contradicting the hypothesis.
next theorem (2.9)