For any quadrilateral the lines containing the diagonals always intersect

Please let me know if the following proof is solid or if it contains errors. Thanks.

Show that for any quadrilateral (convex or not), the lines containing the diagonals always intersect. (From Edwin E. Moise Elementary Geometry from and Advanced Standpoint 3rd Ed.)

Proof: By Theorem 1, the diagonals of a convex quadrilateral always intersect each other. By definition, the diagonals contain two points (the opposite vertices) so by Incidence Axiom 1, the endpoints of the diagonals are contained in one and only one line, therefore the lines containing the diagonals of a convex quadrilateral always intersect.

It is left to show that if the quadrilateral is not convex, the lines containing the diagonals always intersect.

Consider the quadrilateral ABCD. In order for quadrilateral ABCD to not be convex, at least two adjacent points must be on opposite sides of the line formed by the other two adjacent points.

Without loss of generality, let A and D be on opposite sides of the line formed by B and C. By the plane separation theorem, the line containing B and C must intersect $\displaystyle \overline{AD}$ at a point; call this point P. If B-P-C, then we have a contradiction with the definition of quadrilateral ABCD. So we must have B-C-P. (Note that C-B-P leads to the same conclusion.)

This means that C must be in the interior of $\displaystyle \angle BAD $. By the crossbar theorem, any ray $\displaystyle \overrightarrow{AC}$ will intersect the segment $\displaystyle \overline{BD} $. Since the two points A and C are contained in one and only one line by Incedence Axiom 1, the line containing A and C will also intersect segment $\displaystyle \overline{BD}$ which means by Incedence Axiom 1, $\displaystyle \overleftrightarrow{AC}$ will intersect $\displaystyle \overleftrightarrow{BD}. $

Similarly if C-B-P, then B is in the interior of $\displaystyle \angle ADC $, so ray $\displaystyle \overrightarrow{DB}$ will intersect segment $\displaystyle \overline{AC}$ by the crossbar theorem. This leads to $\displaystyle \overleftrightarrow{DB}$ intersecting $\displaystyle \overleftrightarrow{AC}$ by the same arguments.