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

Show that every quadrilateral (convex or not) has a side such that the other two vertices lie on the same side of the line . (From Edwin E. Moise Elementary Geometry from and Advanced Standpoint 3rd Ed.)

Proof: Given quadrilateral WXYZ. We have segments , , , and that intersect at the corners of the quadrilateral.

Now by Incedence Axiom 1, segment has points Z and W which are contained in one and only one line, . By incidence axiom 3 and the definition of a quadrilateral, this line is entirely contained in the plane containing the quadrilateral.

By the plane separation theorem, this line divides the plane containing the quadrilateral into two half-planes.

Now consider the segments and . Since W is on the line, X must lie in one half-plane. Similarly, since Z is on the line, Y must lie in one half-plane. So X and Y must lie in either the same half-plane or opposite half-planes determined by the line .

If X and Y lie in opposite half-planes, then the side of the quadrilateral, , would have to intersect the line by the plane separation theorem. If intersects , this is a contradiction which proves that X amd Y lie in the same half-plane which proves that W and Z lie on the same side of the line .

If intersects the line outside the segment , then Z and W lie in the same half-plane determined by the line .