I came up with the following little question while randomly doodling on a piece of paper.

TheoremGiven a pentagon ABCDE, for a given side, we call the diagonal of the pentagon which does not share a vertex with the side opposite, e.g. the opposite diagonal for the side AB would be the segment CE. If ABCDE has the property that every pair of side and its opposite diagonal are parallel, then ABCDE is, up to an affine transform, the regular pentagon.

As an illustration (for the theorem as well for the proof that follows), we show The condition in the theorem says that if AB is parallel to PQ, BC to QR, CD to RS, DE to ST, and EA to TP, then we can find an affine transform that takes ABCDE to a regular pentagon. (A quick reminder, affine transforms are the linear transforms of the plane plus the translations, so we are allowed to rotate, scale, skew/sear the picture, as well as translate it.)

Proof (The essential step in the proof is due to my office mate Pin Yu.) Since we are proving for equivalence up to affine transforms (note that a fundamental property of affine transforms is that they preserve parallels), we can, without loss of generality, assume that: the lengths AE = ED = DC and the angles AED = EDC. (We can first skew the image parallel to the segment ED via an affine transform such that the angles are as described, then we can take a one dimensional scaling in the direction perpendicular to ED such that the lengths AE = CD are scaled to equal the length of ED.)

It remains to show that every such pentagon is the regular pentagon. To do so:

1. By construction, CDES is a parallelgram, and in fact, a rhombus. Therefore the angles ECD = ECS = CES = CED
2. Since ATDE is a similar rhombus to CSED, the angles ECD = TAD = EAD = EDA = TDA
3. Using the fact that AB is parallel to EC, the angle ABE = BEC, and similarly BAC = ACE, CBD = BDA, BCA = CAD.
4. By parallelisms, we can also say that the angles AEB = EBD = BDC.

Our conclusion is best understood using the following image where the "a" and "b" denotes angle sizes. (Of course, the picture is not to scale, since the angles are only equal after the affine transform mentioned previously.)

So it remains to show that angle b = angle a. (And here's Pin's key insight.)

Look at the triangles ABC and AED, they are similar by the AAA rule. So we must have the ratio

AE / AB = AD / AC

But AE and AB are both sides of the triangle ABE, so by the law of sines

sin a / sin b = AE / AB

Similarly, we get

AD / AC = sin(2a) / sin (a+b)

Working out the math, we have

sin a (sin a cos b + sin b cos a) = 2 sin a cos a sin b

and so

sin a cos b - sin b cos a = sin (a-b) = 0

and hence a = b. Q.E.D.