PROJECTION OF A QUADRANGLE INTO A PARALLELOGRAM1
William H. Richardson
A problem in an old textbook of mine2 reads Show that any quadrangle may be projected into a parallelogram.'' The solution of this problem usually went as follows: Let be the given quadrangle with exterior diagonal points and . Let be any point not in the plane of and join the points with . See Figure 1. Now if the set of lines are cut by a plane parallel to the plane we get points on , on , on and on , which when joined form a parallelogram. The fact that is a parallelogram is established as follows: is on the line of intersection of the planes of and , therefore its image on must be an ideal point since and the plane are parallel. Now would be the intersection of and , so it follows that and are parallel. A similar argument establishes that and are parallel.

What intrigued me about this problem was the fact that although we were dealing with a three dimensional solution to the problem, the two dimensional drawing in Figure 1 made the problem appear solvable in the plane. This led to the investigation of the case in which the point is in the plane of . More precisely, given a quadrangle and a point in its plane, construct a line such that there is a parallelogram which is copolar with from the point and coaxial from the line . See Figure 2.
 Alternate View of not in the plane of

If is in the plane of , the construction is as follows. Join with the vertices and the exterior diagonal points and . On the line pick a point and construct lines and parallel to and , respectively. meets in a point and meets in a point . Join the points and . The line meets at and at . Join the points and which meet at . Join and . This line intersects at and the figure is the required parallelogram.

Proof. Triangles and are coaxial from the line since meets in , meets in and meets in . By Desargues' Theorem, these triangles are copolar. By construction, is parallel to . Therefore, the pole is an ideal point, and thus is parallel to . Since triangles and are coaxial from line , we see, by a similar argument, that is parallel to . All that remains to prove is that and are concurrent at .

Suppose meets at . Triangles and are coaxial from the line , and therefore they must be copolar. Line was constructed parallel to ; therefore is parallel to . On the other hand, it has already been established that is parallel to . Since one and only one line can be drawn parallel to a given line through a given point, is on line . Therefore, lines are concurrent on .

It is worth noting that if is chosen to be on the circle with diameter , the parallelogram will be a rectangle.

Moreover, it appears that there exist points on the circle for which the rectangle becomes a square. Below, in Figure 4, we see rectangles that are produced by choices of close to and close to . In the drawing on the left ; whereas, on the right . As moves from the position on the left to the position on the right, the dimensions of the rectangle change continuously from to . In Figure 5 we see near a point that produces a square.