(To remind, a quadrilateral ABCD is called equilic if it has a pair of equal opposite sides inclined 60° to each other. In the problem, AD = BC.)

Triangles ADQ and BCQ are equal. Indeed, by construction, AD = BC and DQ = CQ. Now, on the surface, the proof proceeds smoothly. We have to consider two possible cases as shown the diagrams below:

We denote the angles in the quadrilateral by a single letter as in the corresponding vertex so that, by construction, A + B = 120°. Therefore, C + D = 240°. Taking up the left diagram:

ADQ = D + 60°   = 240° - C + 60°   = 300° - C   = 360° - C - 60°   = BCQ.

It follows that ΔADQ = ΔBCQ. In particular, AQD = BQC. Thus, by removing and adding equal angles from and to CQD we still get an angle of 60°. Therefore, AQB = 60°. Since also AQ = BQ (as the corresponding sides in equal triangles), ΔABQ is indeed equilateral.

In the case of the right diagram above we similarly arrive at the same conclusion. Q.E.D.

A Flaw in the Proof

A Valid Proof

A rotation around Q through 60° maps D to C and A to, say, A' such that CA' and AD form a 60° angle. But the same is true of AD and CB. Therefore, A' = B. It follows that that rotation maps the whole ΔADQ on top of ΔBCQ. The triangle are then equal, their angles at Q are also equal, and we can proceed as before to conclude that AQB = 60°.