# Equilic Quadrilateral I: What is this about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander BogomolnyThe applet may suggest the following statement due to Jack Garfunkel [Honsberger]:

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,

What if applet does not run? |

Triangles ADQ and BCQ are equal. Indeed, by construction,

We denote the angles in the quadrilateral by a single letter as in the corresponding vertex so that, by construction,

∠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,

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

### Equilic Quadrilateral

- Equilic Quadrilateral I
- Equilic Quadrilateral II
- Equilateral Triangles on Segments of Equilic Quadrilateral
- Equilic Quadrilateral I, A Variation
- Equilateral Triangles on Diagonals of Antiequilic Quadrilateral

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny### 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,

Besides establishing the fact it was intended to prove, the proof has a virtue of suggesting a generalization.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny63231051 |