Bicentric quadrilateral#Fuss' theorem

File:Bicentric kite 001.svg]]

Examples of bicentric quadrilaterals are squares, right kites, and isosceles tangential trapezoids.

File:Bicentric quadrilateral.svg

A convex quadrilateral {{mvar|ABCD}} with sides {{mvar|a, b, c, d}} is bicentric if and only if opposite sides satisfy Pitot's theorem for tangential quadrilaterals and the cyclic quadrilateral property that opposite angles are supplementary; that is,

:$$

\begin{cases}

a+c=b+d\\

A+C=B+D=\pi.

\end{cases}

Three other characterizations concern the points where the incircle in a tangential quadrilateral is tangent to the sides. If the incircle is tangent to the sides {{mvar|AB, BC, CD, DA}} at {{mvar|W, X, Y, Z}} respectively, then a tangential quadrilateral {{mvar|ABCD}} is also cyclic if and only if any one of the following three conditions holds:{{citation

|last=Josefsson

|first=Martin

|journal=Forum Geometricorum

|pages=165–173

|title=Characterizations of Bicentric Quadrilaterals

|url=http://forumgeom.fau.edu/FG2010volume10/FG201019.pdf

|volume=10

|year=2010}}.

• {{mvar|WY}} is perpendicular to {{mvar|XZ}}
• $\backslash frac\{\backslash overline\{AW\}\}\{\backslash overline\{WB\}\}=\backslash frac\{\backslash overline\{DY\}\}\{\backslash overline\{YC\}\}$
• $\backslash frac\{\backslash overline\{AC\}\}\{\backslash overline\{BD\}\}=\backslash frac\{\backslash overline\{AW\}+\backslash overline\{CY\}\}\{\backslash overline\{BX\}+\backslash overline\{DZ\}\}$

The first of these three means that the contact quadrilateral {{mvar|WXYZ}} is an orthodiagonal quadrilateral.

If {{mvar|E, F, G, H}} are the midpoints of {{mvar|WX, XY, YZ, ZW}} respectively, then the tangential quadrilateral {{mvar|ABCD}} is also cyclic if and only if the quadrilateral {{mvar|EFGH}} is a rectangle.

According to another characterization, if {{mvar|I}} is the incenter in a tangential quadrilateral where the extensions of opposite sides intersect at {{mvar|J}} and {{mvar|K}}, then the quadrilateral is also cyclic if and only if {{math|∠ JIK}} is a right angle.

Yet another necessary and sufficient condition is that a tangential quadrilateral {{mvar|ABCD}} is cyclic if and only if its Newton line is perpendicular to the Newton line of its contact quadrilateral {{mvar|WXYZ}}. (The Newton line of a quadrilateral is the line defined by the midpoints of its diagonals.)

File:01-Bicentric quadrilateral.svg]

There is a simple method for constructing a bicentric quadrilateral:

It starts with the incircle {{mvar|C{{sub|r}}}} around the centre {{mvar|I}} with the radius {{mvar|r}} and then draw two to each other perpendicular chords {{mvar|{{overline|WY}}}} and {{mvar|{{overline|XZ}}}} in the incircle {{mvar|C{{sub|r}}}}. At the endpoints of the chords draw the tangents {{mvar|a, b, c, d}} to the incircle. These intersect at four points {{mvar|A, B, C, D}}, which are the vertices of a bicentric quadrilateral.{{cite book

|last1=Alsina

|first1=Claudi

|last2=Nelsen

|first2=Roger

|date=2011

|title=Icons of Mathematics. An exploration of twenty key images

|title-link= Icons of Mathematics

|publisher=Mathematical Association of America

|pages=125–126

|isbn=978-0-88385-352-8}}

To draw the circumcircle, draw two perpendicular bisectors {{math|p{{sub|1}}, p{{sub|2}}}} on the sides of the bicentric quadrilateral {{mvar|a}} respectively {{mvar|b}}. The perpendicular bisectors {{math|p{{sub|1}}, p{{sub|2}}}} intersect in the centre {{mvar|O}} of the circumcircle {{mvar|C{{sub|R}}}} with the distance {{mvar|x}} to the centre {{mvar|I}} of the incircle {{mvar|C{{sub|r}}}}. The circumcircle can be drawn around the centre {{mvar|O}}.

The validity of this construction is due to the characterization that, in a tangential quadrilateral {{mvar|ABCD}}, the contact quadrilateral {{mvar|WXYZ}} has perpendicular diagonals if and only if the tangential quadrilateral is also cyclic.

The area {{mvar|K}} of a bicentric quadrilateral can be expressed in terms of four quantities of the quadrilateral in several different ways. If the sides are {{mvar|a, b, c, d}}, then the area is given by

:$\backslash displaystyle\; K\; =\; \backslash sqrt\{abcd\}.$

This is a special case of Brahmagupta's formula. It can also be derived directly from the trigonometric formula for the area of a tangential quadrilateral. Note that the converse does not hold: Some quadrilaterals that are not bicentric also have area $\backslash displaystyle\; K\; =\; \backslash sqrt\{abcd\}.$Lord, Nick, "Quadrilaterals with area formula $\backslash displaystyle\; K\; =\; \backslash sqrt\{abcd\}.$", Mathematical Gazette 96, July 2012, 345-347. One example of such a quadrilateral is a non-square rectangle.

The area can also be expressed in terms of the tangent lengths {{mvar|e, f, g, h}} as{{rp|p.128}}

:$K=\backslash sqrt[4]\{efgh\}(e+f+g+h).$

A formula for the area of bicentric quadrilateral {{mvar|ABCD}} with incenter {{mvar|I}} is

:$K=\backslash overline\{AI\}\; \backslash cdot\; \backslash overline\{CI\}\; +\; \backslash overline\{BI\}\; \backslash cdot\; \backslash overline\{DI\}.$

If a bicentric quadrilateral has tangency chords {{mvar|k, l}} and diagonals {{mvar|p, q}}, then it has area{{citation

|last=Josefsson

|first=Martin

|journal=Forum Geometricorum

|pages=119–130

|title=Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral

|url=http://forumgeom.fau.edu/FG2010volume10/FG201013.pdf

|volume=10

|year=2010

|access-date=August 19, 2011

|archive-date=August 13, 2011

|archive-url=https://web.archive.org/web/20110813091938/http://forumgeom.fau.edu/FG2010volume10/FG201013.pdf

|url-status=dead

}}.{{rp|p.129}}

:$K=\backslash frac\{klpq\}\{k^2+l^2\}.$

If {{mvar|k, l}} are the tangency chords and {{mvar|m, n}} are the bimedians of the quadrilateral, then the area can be calculated using the formula{{citation

|last=Josefsson

|first=Martin

|journal=Forum Geometricorum

|pages=155–164

|title=The Area of a Bicentric Quadrilateral

|url=http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf

|volume=11

|year=2011

|access-date=October 20, 2011

|archive-date=January 5, 2020

|archive-url=https://web.archive.org/web/20200105031952/http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf

|url-status=dead

}}.

:$K=\backslash left|\backslash frac\{m^2-n^2\}\{k^2-l^2\}\backslash right|kl$

This formula cannot be used if the quadrilateral is a right kite, since the denominator is zero in that case.

If {{mvar|M, N}} are the midpoints of the diagonals, and {{mvar|E, F}} are the intersection points of the extensions of opposite sides, then the area of a bicentric quadrilateral is given by

:$K=\backslash frac\{2\backslash overline\{MN\}\; \backslash cdot\; \backslash overline\{EI\}\; \backslash cdot\; \backslash overline\{FI\}\}\{\backslash overline\{EF\}\}$

where {{mvar|I}} is the center of the incircle.

The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle {{mvar|θ}} between the diagonals according to

:$K=ac\backslash tan\{\backslash frac\{\backslash theta\}\{2\}\}=bd\backslash cot\{\backslash frac\{\backslash theta\}\{2\}\}.$

In terms of two adjacent angles and the radius {{mvar|r}} of the incircle, the area is given by

:$K=2r^2\backslash left(\backslash frac\{1\}\{\backslash sin\{A\}\}+\backslash frac\{1\}\{\backslash sin\{B\}\}\backslash right).$

The area is given in terms of the circumradius {{mvar|R}} and the inradius {{mvar|r}} as

:$K=r(r+\backslash sqrt\{4R^2+r^2\})\backslash sin\; \backslash theta$

where {{mvar|θ}} is either angle between the diagonals.{{citation

|last=Josefsson

|first=Martin

|journal=Forum Geometricorum

|pages=237–241

|title=Maximal Area of a Bicentric Quadrilateral

|url=http://forumgeom.fau.edu/FG2012volume12/FG201222.pdf

|volume=12

|year=2012

|access-date=October 23, 2012

|archive-date=December 5, 2022

|archive-url=https://web.archive.org/web/20221205165012/https://forumgeom.fau.edu/FG2012volume12/FG201222.pdf

|url-status=dead

}}.

If {{mvar|M, N}} are the midpoints of the diagonals, and {{mvar|E, F}} are the intersection points of the extensions of opposite sides, then the area can also be expressed as

:$K=2\backslash overline\{MN\}\backslash sqrt\{\backslash overline\{EQ\}\; \backslash cdot\; \backslash overline\{FQ\}\}$

where {{mvar|Q}} is the foot of the perpendicular to the line {{mvar|EF}} through the center of the incircle.

If {{mvar|r}} and {{mvar|R}} are the inradius and the circumradius respectively, then the area {{mvar|K}} satisfies the inequalities{{cite book

|last1=Alsina

|first1=Claudi

|last2=Nelsen

|first2=Roger

|date=2009

|title=When less is more: visualizing basic inequalities

|url=https://archive.org/details/whenlessismorevi00alsi_112

|url-access=limited

|publisher=Mathematical Association of America

|pages=[https://archive.org/details/whenlessismorevi00alsi_112/page/n84 64]–66

|isbn=978-0-88385-342-9}}

:$\backslash displaystyle\; 4r^2\; \backslash le\; K\; \backslash le\; 2R^2.$

There is equality on either side only if the quadrilateral is a square.

Another inequality for the area isInequalities proposed in Crux Mathematicorum, 2007.[http://hydra.nat.uni-magdeburg.de/math4u/ineq.pdf]{{rp|p.39,#1203}}

:$K\; \backslash le\; \backslash tfrac\{4\}\{3\}r\backslash sqrt\{4R^2+r^2\}$

where {{mvar|r}} and {{mvar|R}} are the inradius and the circumradius respectively.

A similar inequality giving a sharper upper bound for the area than the previous one is

:$K\; \backslash le\; r(r+\backslash sqrt\{4R^2+r^2\})$

with equality holding if and only if the quadrilateral is a right kite.

In addition, with sides {{mvar|a, b, c, d}} and semiperimeter {{mvar|s}}:

:$2\backslash sqrt\{K\}\; \backslash leq\; s\; \backslash leq\; r+\; \backslash sqrt\{r^2+4R^2\};${{rp|p.39,#1203}}

:$6K\; \backslash leq\; ab+ac+ad+bc+bd+cd\; \backslash leq\; 4r^2+4R^2+\; 4r\backslash sqrt\{r^2+4R^2\};${{rp|p.39,#1203}}

:$4Kr^2\backslash leq\; abcd\; \backslash leq\; \backslash frac\{16\}\{9\}\; r^2(r^2+4R^2).${{rp|p.39,#1203}}

If {{mvar|a, b, c, d}} are the length of the sides {{mvar|AB, BC, CD, DA}} respectively in a bicentric quadrilateral {{mvar|ABCD}}, then its vertex angles can be calculated with the tangent function:

:$\backslash begin\{align\}$

\tan{\frac{A}{2}} &= \sqrt{\frac{bc}{ad}} = \cot{\frac{C}{2}}, \\

\tan{\frac{B}{2}} &= \sqrt{\frac{cd}{ab}} = \cot{\frac{D}{2}}.

\end{align}

Using the same notations, for the sine and cosine functions the following formulas holds:{{citation

|last=Josefsson

|first=Martin

|journal=Forum Geometricorum

|pages=79–82

|title=A New Proof of Yun's Inequality for Bicentric Quadrilaterals

|url=http://forumgeom.fau.edu/FG2012volume12/FG201208.pdf

|volume=12

|year=2012

|access-date=June 23, 2012

|archive-date=December 31, 2019

|archive-url=https://web.archive.org/web/20191231065908/http://forumgeom.fau.edu/FG2012volume12/FG201208.pdf

|url-status=dead

}}.

:$\backslash begin\{align\}$

\sin{\frac{A}{2}} &= \sqrt{\frac{bc}{ad+bc}} = \cos{\frac{C}{2}}, \\

\cos{\frac{A}{2}} &= \sqrt{\frac{ad}{ad+bc}} = \sin{\frac{C}{2}}, \\

\sin{\frac{B}{2}} &= \sqrt{\frac{cd}{ab+cd}} = \cos{\frac{D}{2}}, \\

\cos{\frac{B}{2}} &= \sqrt{\frac{ab}{ab+cd}} = \sin{\frac{D}{2}}.

\end{align}

The angle {{mvar|θ}} between the diagonals can be calculated fromDurell, C. V. and Robson, A., Advanced Trigonometry, Dover, 2003, pp. 28, 30.

:$\backslash displaystyle\; \backslash tan\{\backslash frac\{\backslash theta\}\{2\}\}=\backslash sqrt\{\backslash frac\{bd\}\{ac\}\}.$

The inradius {{mvar|r}} of a bicentric quadrilateral is determined by the sides {{mvar|a, b, c, d}} according toWeisstein, Eric, Bicentric Quadrilateral at MathWorld, [http://mathworld.wolfram.com/BicentricQuadrilateral.html], Accessed on 2011-08-13.

:$\backslash displaystyle\; r=\backslash frac\{\backslash sqrt\{abcd\}\}\{a+c\}=\backslash frac\{\backslash sqrt\{abcd\}\}\{b+d\}.$

The circumradius {{mvar|R}} is given as a special case of Parameshvara's formula. It is

:$\backslash displaystyle\; R=\backslash frac\{1\}\{4\}\backslash sqrt\{\backslash frac\{(ab+cd)(ac+bd)(ad+bc)\}\{abcd\}\}.$

The inradius can also be expressed in terms of the consecutive tangent lengths {{mvar|e, f, g, h}} according toM. Radic, Z. Kaliman, and V. Kadum, "A condition that a tangential quadrilateral is also a chordal one", Mathematical Communications, 12 (2007) 33–52.{{rp|p. 41}}

:$\backslash displaystyle\; r=\backslash sqrt\{eg\}=\backslash sqrt\{fh\}.$

These two formulas are in fact necessary and sufficient conditions for a tangential quadrilateral with inradius {{mvar|r}} to be cyclic.

The four sides {{mvar|a, b, c, d}} of a bicentric quadrilateral are the four solutions of the quartic equation

:$y^4-2sy^3+(s^2+2r^2+2r\backslash sqrt\{4R^2+r^2\})y^2-2rs(\backslash sqrt\{4R^2+r^2\}+r)y+r^2s^2=0$

where {{mvar|s}} is the semiperimeter, and {{mvar|r}} and {{mvar|R}} are the inradius and circumradius respectively.Pop, Ovidiu T., "Identities and inequalities in a quadrilateral", Octogon Mathematical Magazine, Vol. 17, No. 2, October 2009, pp 754-763.{{rp|p. 754}}

If there is a bicentric quadrilateral with inradius {{mvar|r}} whose tangent lengths are {{mvar|e, f, g, h}}, then there exists a bicentric quadrilateral with inradius {{mvar|r{{sup|v}}}} whose tangent lengths are {{tmath|e^v,f^v,g^v,h^v,}} where {{mvar|v}} may be any real number.{{rp|pp.9–10}}

A bicentric quadrilateral has a greater inradius than does any other tangential quadrilateral having the same sequence of side lengths.{{citation

|last=Hess

|first=Albrecht

|journal=Forum Geometricorum

|pages=389–396

|title=On a circle containing the incenters of tangential quadrilaterals

|url=http://forumgeom.fau.edu/FG2014volume14/FG201437.pdf

|volume=14

|year=2014

|access-date=December 15, 2014

|archive-date=December 14, 2014

|archive-url=https://web.archive.org/web/20141214205151/http://forumgeom.fau.edu/FG2014volume14/FG201437.pdf

|url-status=dead

}}.{{rp|pp.392–393}}

The circumradius {{mvar|R}} and the inradius {{mvar|r}} satisfy the inequality

:$R\backslash ge\; \backslash sqrt\{2\}r$

which was proved by L. Fejes Tóth in 1948.Radic, Mirko, "Certain inequalities concerning bicentric quadrilaterals, hexagons and octagons", Journal of Inequalities in Pure and Applied Mathematics, Volume 6, Issue 1, 2005, [http://www.emis.de/journals/JIPAM/images/118_04_JIPAM/118_04.pdf] It holds with equality only when the two circles are concentric (have the same center as each other); then the quadrilateral is a square. The inequality can be proved in several different ways, one using the double inequality for the area above.

An extension of the previous inequality isYun, Zhang, "Euler's Inequality Revisited", Mathematical Spectrum, Volume 40, Number 3 (May 2008), pp. 119-121. First page available at [http://ms.appliedprobability.org/data/files/Abstracts%2040/40-3-6.pdf] {{Webarchive|url=https://web.archive.org/web/20160304085301/http://ms.appliedprobability.org/data/files/Abstracts%2040/40-3-6.pdf |date=March 4, 2016 }}.Shattuck, Mark, “A Geometric Inequality for Cyclic Quadrilaterals”, Forum Geometricorum 18, 2018, 141-154. [http://forumgeom.fau.edu/FG2018volume18/FG201822.pdf] {{Webarchive|url=https://web.archive.org/web/20180913223454/http://forumgeom.fau.edu/FG2018volume18/FG201822.pdf |date=September 13, 2018 }} This paper also gives various inequalities in terms of the arc lengths subtended by a cyclic quadrilateral’s sides.{{rp|p. 141}}

:$\backslash frac\{r\backslash sqrt\{2\}\}\{R\}\backslash le\; \backslash frac\{1\}\{2\}\backslash left(\backslash sin\{\backslash frac\{A\}\{2\}\}\backslash cos\{\backslash frac\{B\}\{2\}\}+\backslash sin\{\backslash frac\{B\}\{2\}\}\backslash cos\{\backslash frac\{C\}\{2\}\}+\backslash sin\{\backslash frac\{C\}\{2\}\}\backslash cos\{\backslash frac\{D\}\{2\}\}+\backslash sin\{\backslash frac\{D\}\{2\}\}\backslash cos\{\backslash frac\{A\}\{2\}\}\backslash right)\backslash le\; 1$

where there is equality on either side if and only if the quadrilateral is a square.{{rp|p. 81}}

The semiperimeter {{mvar|s}} of a bicentric quadrilateral satisfies{{rp|p.13}}

:$\backslash sqrt\{8r\backslash left(\backslash sqrt\{4R^2+r^2\}-r\backslash right)\}\backslash le\; s\; \backslash le\; \backslash sqrt\{4R^2+r^2\}+r$

where {{mvar|r}} and {{mvar|R}} are the inradius and circumradius respectively.

Moreover,{{rp|p.39,#1203}}

:$2sr^2\backslash leq\; abc+abd+acd+bcd\; \backslash leq\; 2r(r+\backslash sqrt\{r^2+4R^2\})^2$

and

:$abc+abd+acd+bcd\; \backslash leq\; 2\backslash sqrt\{K\}(K+2R^2).$ {{rp|p.62,#1599}}

File:Fuss theorem2.svg

Fuss's theorem gives a relation between the inradius {{mvar|r}}, the circumradius {{mvar|R}} and the distance {{mvar|x}} between the incenter {{mvar|I}} and the circumcenter {{mvar|O}}, for any bicentric quadrilateral. The relation isYiu, Paul, Euclidean Geometry, [http://math.fau.edu/Yiu/EuclideanGeometryNotes.pdf] {{Webarchive|url=https://web.archive.org/web/20190302221424/http://math.fau.edu/Yiu/EuclideanGeometryNotes.pdf |date=March 2, 2019 }}, 1998, pp. 158-164.{{citation |last=Salazar |first=Juan Carlos |title=Fuss's Theorem |journal=Mathematical Gazette |volume=90 (July) |pages=306–307 |year=2006|doi=10.1017/S002555720017980X }}.

:$\backslash frac\{1\}\{(R-x)^2\}+\backslash frac\{1\}\{(R+x)^2\}=\backslash frac\{1\}\{r^2\},$

or equivalently

:$\backslash displaystyle\; 2r^2(R^2+x^2)=(R^2-x^2)^2.$

It was derived by Nicolaus Fuss (1755–1826) in 1792. Solving for {{mvar|x}} yields

:$x=\backslash sqrt\{R^2+r^2-r\backslash sqrt\{4R^2+r^2\}\}.$

Fuss's theorem, which is the analog of Euler's theorem for triangles for bicentric quadrilaterals, says that if a quadrilateral is bicentric, then its two associated circles are related according to the above equations. In fact the converse also holds: given two circles (one within the other) with radii {{mvar|R}} and {{mvar|r}} and distance {{mvar|x}} between their centers satisfying the condition in Fuss's theorem, there exists a convex quadrilateral inscribed in one of them and tangent to the other{{citation |last=Byerly |first=W. E. |title=The In- and-Circumscribed Quadrilateral |journal=The Annals of Mathematics |volume=10 |pages=123–128 |year=1909 |issue=3 |doi=10.2307/1967103|jstor=1967103 }}. (and then by Poncelet's closure theorem, there exist infinitely many of them).

Applying $x^2\; \backslash ge\; 0$ to the expression of Fuss's theorem for {{mvar|x}} in terms of {{mvar|r}} and {{mvar|R}} is another way to obtain the above-mentioned inequality $R\; \backslash ge\; \backslash sqrt\{2\}r.$ A generalization is{{rp|p.5}}

:$2r^2+x^2\backslash le\; R^2\; \backslash le\; 2r^2+x^2+2rx.$

Another formula for the distance {{mvar|x}} between the centers of the incircle and the circumcircle is due to the American mathematician Leonard Carlitz (1907–1999). It states thatCalin, Ovidiu, Euclidean and Non-Euclidean Geometry a metric approach, [http://math.emich.edu/~ocalin/Math341/Newpdf/driver13GeomB.pdf], pp. 153–158.

:$\backslash displaystyle\; x^2=R^2-2Rr\backslash cdot\; \backslash mu$

where {{mvar|r}} and {{mvar|R}} are the inradius and the circumradius respectively, and

:$\backslash displaystyle\; \backslash mu=\backslash sqrt\{\backslash frac\{(ab+cd)(ad+bc)\}\{(a+c)^2(ac+bd)\}\}\; =\; \backslash sqrt\{\backslash frac\{(ab+cd)(ad+bc)\}\{(b+d)^2(ac+bd)\}\}$

where {{mvar|a, b, c, d}} are the sides of the bicentric quadrilateral.

For the tangent lengths {{mvar|e, f, g, h}} the following inequalities holds:{{rp|p.3}}

:$4r\backslash le\; e+f+g+h\; \backslash le\; 4r\backslash cdot\; \backslash frac\{R^2+x^2\}\{R^2-x^2\}$

and

:$4r^2\backslash le\; e^2+f^2+g^2+h^2\; \backslash le\; 4(R^2+x^2-r^2)$

where {{mvar|r}} is the inradius, {{mvar|R}} is the circumradius, and {{mvar|x}} is the distance between the incenter and circumcenter. The sides {{mvar|a, b, c, d}} satisfy the inequalities{{rp|p.5}}

:$8r\backslash le\; a+b+c+d\; \backslash le\; 8r\backslash cdot\; \backslash frac\{R^2+x^2\}\{R^2-x^2\}$

and

:$4(R^2-x^2+2r^2)\backslash le\; a^2+b^2+c^2+d^2\; \backslash le\; 4(3R^2-2r^2).$

The circumcenter, the incenter, and the intersection of the diagonals in a bicentric quadrilateral are collinear.Bogomolny, Alex, Collinearity in Bicentric Quadrilaterals [https://www.cut-the-knot.org/ctk/BicentricQuadri.shtml], 2004.

There is the following equality relating the four distances between the incenter {{mvar|I}} and the vertices of a bicentric quadrilateral {{mvar|ABCD}}:L. V. Nagarajan, Bi-centric Polygons, 2014, [https://lvnaga.wordpress.com/2014/05/13/bi-centric-polygons/].

:$\backslash frac\{1\}\{\backslash overline\{AI\}^2\}+\backslash frac\{1\}\{\backslash overline\{CI\}^2\}=\backslash frac\{1\}\{\backslash overline\{BI\}^2\}+\backslash frac\{1\}\{\backslash overline\{DI\}^2\}=\backslash frac\{1\}\{r^2\}$

where {{mvar|r}} is the inradius.

If {{mvar|P}} is the intersection of the diagonals in a bicentric quadrilateral {{mvar|ABCD}} with incenter {{mvar|I}}, thenCrux Mathematicorum 34 (2008) no 4, p. 242.

:$\backslash frac\{\backslash overline\{AP\}\}\{\backslash overline\{CP\}\}=\backslash frac\{\backslash overline\{AI\}^2\}\{\backslash overline\{CI\}^2\}.$

The lengths of the diagonals in a bicentric quadrilateral can be expressed in terms of the sides or the tangent lengths, which are formulas that holds in a cyclic quadrilateral and a tangential quadrilateral respectively.

In a bicentric quadrilateral with diagonals {{mvar|p, q}}, the following identity holds:

:$\backslash displaystyle\; \backslash frac\{pq\}\{4r^2\}-\backslash frac\{4R^2\}\{pq\}=1$

where {{mvar|r}} and {{mvar|R}} are the inradius and the circumradius respectively. This equality can be rewritten as

:$r=\backslash frac\{pq\}\{2\backslash sqrt\{pq+4R^2\}\}$

or, solving it as a quadratic equation for the product of the diagonals, in the form

:$pq=2r\backslash left(r+\backslash sqrt\{4R^2+r^2\}\backslash right).$

An inequality for the product of the diagonals {{mvar|p, q}} in a bicentric quadrilateral is

:$\backslash displaystyle\; 8pq\backslash le\; (a+b+c+d)^2$

where {{mvar|a, b, c, d}} are the sides. This was proved by Murray S. Klamkin in 1967.

Let {{mvar|ABCD}} be a bicentric quadrilateral and {{mvar|O}} the center of its circumcircle. Then the incenters of the four triangles {{math|△OAB, △OBC, △OCD, △ODA}} lie on a circle.Alexey A. Zaslavsky, One property of bicentral quadrilaterals, 2019, [http://geometry.ru/olimp/2019/bicentr.pdf]

{{Commons category|Bicentric quadrilateral}}

• Bicentric polygon
• Ex-tangential quadrilateral

{{reflist}}

{{Polygons}}

Category:Types of quadrilaterals