# Easy Construction of Bicentric Quadrilateral II:

What Is This About?

A Mathematical Droodle

What if applet does not run? |

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Copyright © 1996-2018 Alexander Bogomolny

### Easy Construction of Bicentric Quadrilateral

A quadrilateral is *bicentric* if it's both *inscriptable* and *circumscriptable*. (*Inscriptible* means admitting an incircle. *Circumscriptible* means cyclic, i.e., admitting a circumcircle.) Bicentric quadrilaterals might seem exotic, but the applet shows how such quadrilaterals can be constructed easily. (Another construction appears elsewhere.)

What if applet does not run? |

Let ABCD be a cyclic quadrilateral with vertices on a given circle w. Let P denote the point of intersection of the diagonals AC and BD. From P drop the perpendiculars PK, PL, ... to the sides, with K on AB, and so on. Our first assertion is that the quadrilateral KLMN is inscriptible. (As the proof shows, if we start with an arbitrary quadrlateral ABCD, then KLMN is inscriptible iff ABCD is cyclic.)

### Propostition 1

KLMN is inscriptible.

### Proof

By construction, the quadrilateral KBLP is cyclic. Therefore,

∠KLP = ∠KBP, |

as subtending the same chord KP. Note that ∠KBP = ∠ABD. Since LCMP is also cyclic, we similarly obtain

∠MLP = ∠MCP = ∠BCD. |

For a cyclic quadrilateral ABCD, ∠ABD = ∠ACD, so that

For KLMN to be bicentric, it must be made circumscriptible - cyclic. The condition for that is given by

### Proposition 2

KLMN is cyclic iff ABCD is orthodiagonal.

### Proof

Assume ABCD is *orthodiagonal*, i.e.,

Rectangle is a cyclic shape with the circumcenter at the intersection of the diagonals, both of which are diameters of the incircle q. Now, for example, in ΔLL'N', angle L is right, while the hypotenuse L'N' is a diameter of circle q. L therefore lies on q. The same of course holds for K, M, and N.

To sum up, quadrilateral KLMN is inscribed in a circle centered at the intersection O of K'M' and L'N'.

Conversely, assume KLMN is cyclic and let's evaluate the angles in ΔAPB. We'll show that

First of all, the cyclicity of KLMN is equivalent to

(1) | ∠LKN + ∠LMN = 180°. |

By construction, four quadrilaterals MPND, MPLC, LPKB, and KPNA are also cyclic. From this we successively obtain

(2) |
∠NDP = ∠NMP, ∠LCP = ∠LMP, ∠LBP = ∠LKP, ∠NAP = ∠NKP. |

Obviously

(3) |
∠NDP + ∠NAP = ∠CPD, ∠LBP + ∠LCP = ∠APB, ∠NMP + ∠LMP = ∠LMN, ∠LKP + ∠NKP = ∠LKN. |

Together (2) and (3) lead to

(4) | ∠LKN + ∠LMN = ∠APB + ∠CPD = 2∠APB. |

The required ∠APB = 90° now follows from (4) and (1).

The applet also suggests that the three points O, P, and E are collinear. This is true and could be shown in a very general framework.

### References

- R. Honsberger,
*In Pólya's Footsteps*, MAA, 1999, pp 60-64 - A. A. Zaslavsky,
__The Orthodiagonal Mapping of Quadrilaterals__,*Kvant*, n 4, 1998, pp 43-44 (in Russian), pdf is available at https://kvant.mccme.ru/1998/04/.

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Copyright © 1996-2018 Alexander Bogomolny