Theorem 2.13: (The first Pasch property) Let   A,   B,   and   C   be three noncolinear points. If a line going through   A   contains points in the angle between   AB   and   BC,   then the line intersects the line segment   BC.

Proof: Suppose a line through   A   contains a point   D   which is inside the angle.

The illustration indicates that   D   can be on either side of the line determined by   B   and   C,   so long as it is inside the angle between   AB   and   BC.

Then by Theorem 2.11, there are real numbers   q,   r,   and   s   such that

D = qA + rB + sC

where

q + r + s = 1.

Since by the definition of the angle,   D   is on the same side of   AB   as   C,   it follows from Theorem 2.12 that   s   is positive. Similarly, we can conclude that   r   is also positive. Since   r   and   s   are both positive we can define

Then by Theorem 2.1

E = (1 - t)A + tD

is on the line determined by   A   and   D.   But

E = (1 - t)A + tD

Find common denominators in the coefficient of the first term, and remove the parentheses in the second term.

Combine the   A   terms

But q + r + s = 1,   so the   A  term has a zero coefficient. We get

However,

so   E   is on the line determined by   B   and   C   by Theorem 2.1. Moreover, since   r   and   s   are both positive   r/(r + s)   and   s/(r + s)   are both between   0   and   1,   so by the definition of the line segment between two points,   E  is on the line segment between   B   and   C.

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