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Equilic Quadrilateral I: What is this about?
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The applet may suggest the following statement due to Jack Garfunkel [Honsberger]:
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If an equilateral triangle DCQ is drawn on the side DC (away from AB) of an equilic quadrilateral ABCD |
(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,
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Triangles ADQ and BCQ are equal. Indeed, by construction,
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We denote the angles in the quadrilateral by a single letter as in the corresponding vertex so that, by construction, 
A + B = 120°.C + D = 240°.
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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°.
In the case of the right diagram above we similarly arrive at the same conclusion. Q.E.D.
A Flaw in the Proof
Now, this proof contains a subtle flaw. For the convenience sake, to simplify the process of drawing an equilic quadrilateral, I used the fact that the two sides AD and BC make a 60° angle so that their intersection S is bound to a 240° arc above AB. The statement we have attempted to prove is equivalent to stating that Q lies at the middle of that arc. The argument was based on the fact that their were just two possibilities: either C or D is inside ΔABQ whereas the other one is outside. However, if Q does not lie on the arc, two other possibilities have to be considered: either both C and D are inside or outside ΔABQ. It appears that we based our proof on an assumption equivalent to the statement to be proved.
References
- R. Honsberger, Mathematical Gems III, MAA, 1985, pp. 32-35
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, AQB = 60°.
Besides establishing the fact it was intended to prove, the proof has a virtue of suggesting a generalization.
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