Easy Construction of Bicentric Quadrilateral II:
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Copyright © 1996-2012 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.)
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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 http://kvant.mccme.ru/1998/04/.
Bicentric Quadrilateral
- Collinearity in Bicentric Quadrilaterals
- Easy Construction of Bicentric Quadrilateral
- Easy Construction of Bicentric Quadrilateral II
- Projective Collinearity in a Quadrilateral
- Fuss' Theorem
- Line IO in Bicentric Quadrilaterals
- Area of a Bicentric Quadrilateral
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