**Theorem 2.14**: (The second
Pasch property) Let * A*, * * *B*, * * and * * *C* * * be three noncolinear points. If a
line crosses the
line segment * AB*, * * then the line will either intersect
line segment * * *AC*, * * segment * * *BC*, * * or go through point * **C*.

**Proof**: Let * * *D* * * be the point where the
line crosses line
segment * * *AB*. * * We can conclude from Theorem 2.1 that there are
points on the
line segment between * A* * * and * * *B*, * *which are on either side of the
point * * *D*. * * By Theorem 2.5, then, there are
points on that
line segment which are on either
sides of the
line determined by * * *A* * * and * * *B*.
* * Let * * *E* * * be a point on the
same side of the
line determined by * * *AB* * * as * * *C*.

Consider the line determined by * * *CD*. * * Since * * *C* * * is, by hypotheses, not on the line determined by * * *AB*, * * we
conclude that * CD* * * crosses * * *AB* * * at * * *D*. * * As a result, by Theorem 2.5, * * *A* * * is on one side of the
line determined by * * *CD*, * * and * * *B* * * is on the other side.

If we substitute the coordinates of * E* * * into the equation for the line determined by * CD*, * *the
trichotomy law for real numbers states that there are three cases.

**Case 1**: * * *E* * * is on * * *CD*. * * Then the line determined by * * *CD* * * and
the line determined by * * *DE*
* * both contain the distinct points * D* * * and * E*, * * so by Theorem 1.4, they are the
same line, and the
line determined by * * *DE*
* * contains the point * * *C*.

**Case 2**: * E *is on the same side of * CD* * * as * * *A*. * * We
also have that * E *is on the same
side of * * *AD* * * as * * *C*, * * because * * *E* * * is supposed to be on the same side of * * *AB* * * as * * *C*, * * and * * *AD* * * has the same equation as * * *AB,* * * so * * *E* * * is on the same side of * AD* * * as * * *C*. * * The
hypotheses of Theorem 2.13 are satisfied,
so we can conclude that * * *DE* * * crosses the line segment between * A* * * and * * *C*.

**Case 3**: * * *E* * * is on the same side of * * *CD* * * as * * *B*. * * This
case is the same as Case 2 if we simply switch * * *A*'s * * and * * *B*'s, * * and the
conclusion is that * * *DE* * * crosses the line segment between * * *B* * *and * * *C*.