### Squares on a Quadrilateral

Tuesday, September 1st, 2020A recent **Quora** question asks the following:

Squares are constructed externally on the sides of an arbitrary quadrilateral. How do you show that the line segments joining the centers of opposite squares lie on perpendicular lines and are of equal length?

I chose to answer it because it provides a nice opportunity both to use what I like to call “continuous induction” and to illustrate the proof with an interactive diagram.

The result is clearly true for a square. So it will suffice to show that if we have any quadrilateral for which it is true and move just one corner then it will be true for the new quadrilateral. A picture would help, but I think you will enjoy (and learn more from) the effort of making your own rather than just scrolling down to look at mine.

I found it helpful to use vector language for describing the points involved, but there is also a nice geometric version of the argument that you might enjoy looking for before scrolling down.

Let the quadrilateral be ABCD and consider the effect of moving B by some displacement vector v. Let’s use s for the side AB and t for the side BC. Then the centre of the square on AB is displaced from A by the vector (s+Rs)/2 where R is a rotation by Pi/2 (counterclockwise if ABCD represents a clockwise orientation of the corners of the quadrilateral), and the centre of the square on BC is displaced from A by s+(t+Rt)/2.

After moving B by the vector v, the vector s is replaced by s+v and t by t-v.

So the centre of the AB square is moved by ((s+v)+R(s+v))/2 -(s+Rs)/2=(v+Rv)/2

and for the BC square it’s moved by [s+v+{(t-v)+R(t-v)}/2]-[s+(t+Rt)/2]=(v-Rv)/2

But since R^2=-1 (two right angle rotations give a reversal), R(v-Rv)/2=(v+Rv)/2

So the motions of the centres are perpendicular and of equal length (and the direction of the rotation from one displacement to the other is the same as that from one midpoint line to the other)

But if the lines joining the opposite pairs of centres are perpendicular and of equal length, and the displacements of the new centres are also perpendicular and of equal length, then (with the directions of rotation matching) the triangles formed by old and new centre lines and their endpoint displacements are congruent and so the lengths of the new centre lines are also of equal length and the angle between them is the same as between the old ones.

If you drew a picture to follow along with the above it probably looked something like this:

But I’ve now made a better version:

which also has the advantage of showing a more intuitive and geometrical version of the argument. So the promised shorter proof goes as follows:

When a corner B is moved to a new position B1, the lines joining its counterclockwise neighbour A to the old and new centres S and S1 are rotated counterclockwise by 45 degrees from the corresponding old and new edges, and scaled by a factor of 1/root(2). So the displacement SS1 is similarly related to BB1. And by the same argument the displacement TT1 of the centre of the square on the neighbouring side in the other direction is also scaled by a factor of 1/root(2) and rotated by 45 degrees in the clockwise direction. So SS1 and TT1 are perpendicular and of equal length. So if we start with a case where the midpoint-joining lines US and VT are perpendicular and of equal length, then by sAs congruence we see that US1 and VT1 are also equal and perpendicular. But, as noted at the outset, the result is obvious for a square and any other quadrilateral can be reached from a square by just moving vertices one-by-one.