(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:
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 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°.
Besides establishing the fact it was intended to prove, the proof has a virtue of suggesting a generalization.
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Copyright © 1996-2012 Alexander Bogomolny
