Every quadrilateral has a side such that the other two vertices lie on the same side

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 $\overline{XY}$ such that the other two vertices lie on the same side of the line $\overleftrightarrow{XY}$ . (From Edwin E. Moise Elementary Geometry from and Advanced Standpoint 3rd Ed.)

Proof: Given quadrilateral WXYZ. We have segments $\overline{WX}$ , $\overline{XY}$ , $\overline{YZ}$ , and $\overline{ZW}$ that intersect at the corners of the quadrilateral.

Now by Incedence Axiom 1, segment $\overline{ZW}$ has points Z and W which are contained in one and only one line, $\overleftrightarrow{ZW}$ . 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 $\overline{WX}$ and $\overline{YZ}$ . 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 $\overleftrightarrow{ZW}$ .

If X and Y lie in opposite half-planes, then the side of the quadrilateral, $\overline{XY}$ , would have to intersect the line $\overleftrightarrow{ZW}$ by the plane separation theorem. If $\overline{XY}$ intersects $\overline{ZW}$ , 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 $\overleftrightarrow{XY}$ .

If $\overline{XY}$ intersects the line $\overleftrightarrow{ZW}$ outside the segment $\overline{ZW}$ , then Z and W lie in the same half-plane determined by the line $\overleftrightarrow{XY}$ .