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| [[Image:Tangential quadrilateral.svg|thumb|An example of a tangential quadrilateral]]
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| In [[Euclidean geometry]], a '''tangential quadrilateral''' (sometimes just '''tangent quadrilateral''') or '''circumscribed quadrilateral''' is a [[convex polygon|convex]] [[quadrilateral]] whose sides are all [[tangent]] to a single [[circle]] within the quadrilateral. This circle is called the [[Incircle and excircles of a triangle|incircle]] of the quadrilateral or its inscribed circle, its center is the ''incenter'' and its radius is called the ''inradius''. Since these quadrilaterals can be drawn surrounding or circumscribing their incircles, they have also been called ''circumscribable quadrilaterals'', ''circumscribing quadrilaterals'', and ''circumscriptible quadrilaterals''.<ref name=Josefsson2/> Tangential quadrilaterals are a special case of [[tangential polygon]]s.
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| Other, rarely used, names for this class of quadrilaterals are ''inscriptable quadrilateral'', ''inscriptible quadrilateral'', ''inscribable quadrilateral'', ''circumcyclic quadrilateral'', and ''co-cyclic quadrilateral''.<ref name=Josefsson2/><ref name=Bryant>Bryant, Victor and Duncan, John, "Wheels within wheels", ''[[Mathematical Gazette]]'' 94, November 2010, 502–505.</ref> Due to the risk of confusion with a quadrilateral that has a circumcircle, which is called a [[cyclic quadrilateral]] or inscribed quadrilateral, it is preferable not to use any of the last five names.<ref name=Josefsson2/>
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| All [[triangle]]s have an incircle, but not all quadrilaterals do. An example of a quadrilateral that cannot be tangential is a non-square [[rectangle]]. The section [[Tangential quadrilateral#Characterizations|characterizations]] below states what [[necessary and sufficient condition]]s a quadrilateral must satisfy to have an incircle.
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| ==Special cases==
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| Examples of tangential quadrilaterals are [[square (geometry)|squares]], [[rhombus|rhombi]], and [[kite (geometry)|kites]]. The kites are exactly the tangential quadrilaterals that are also [[orthodiagonal quadrilateral|orthodiagonal]].<ref name=Josefsson>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=119–130
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| |title=Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral
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| |url=http://forumgeom.fau.edu/FG2010volume10/FG201013.pdf
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| |volume=10
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| |year=2010}}.</ref> If a quadrilateral is both tangential and [[cyclic quadrilateral|cyclic]], it is called a [[bicentric quadrilateral]].
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| ==Characterizations==
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| In a tangential quadrilateral, the four [[angle bisector]]s meet at the center of the incircle. Conversely, a convex quadrilateral in which the four angle bisectors meet at a point must be tangential and the common point is the incenter.<ref name=Andreescu/>
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| According to the [[Pitot theorem]], the two pairs of opposite sides in a tangential quadrilateral add up to the same total length, which equals the [[semiperimeter]] ''s'' of the quadrilateral:
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| :<math>a + c = b + d = \frac{a + b + c + d}{2} = s.</math>
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| Conversely a convex quadrilateral in which ''a'' + ''c'' = ''b'' + ''d'' must be tangential.<ref name=Josefsson2/>{{rp|p.65}}<ref name=Andreescu/>
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| If opposite sides in a convex quadrilateral ''ABCD'' (that is not a [[trapezoid]]) intersect at ''E'' and ''F'', then it is tangential [[if and only if]] either of<ref name=Andreescu>Andreescu, Titu and Enescu, Bogdan, ''Mathematical Olympiad Treasures'', Birkhäuser, 2006, pp. 64–68.</ref>
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| :<math>\displaystyle BE+BF=DE+DF</math>
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| or
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| :<math>\displaystyle AE-EC=AF-FC.</math>
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| The second of these is almost the same as one of the equalities in [[Ex-tangential quadrilateral#Urquhart's theorem|Urquhart's theorem]]. The only differences are the signs on both sides; in Urquhart's theorem there are sums instead of differences. | |
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| Another necessary and sufficient condition is that a convex quadrilateral ''ABCD'' is tangential if and only if the incircles in the two triangles ''ABC'' and ''ADC'' are [[tangent]] to each other.<ref name=Josefsson2>{{citation
| |
| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=65–82
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| |title=More Characterizations of Tangential Quadrilaterals
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| |url=http://forumgeom.fau.edu/FG2011volume11/FG201108.pdf
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| |volume=11
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| |year=2011}}.</ref>{{rp|p.66}}
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| A characterization regarding the angles formed by diagonal ''BD'' and the four sides of a quadrilateral ''ABCD'' is due to Iosifescu. He proved in 1954 that a convex quadrilateral has an incircle if and only if<ref name=Minculete>
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| {{citation
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| |last=Minculete |first=Nicusor
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| |journal=Forum Geometricorum
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| |pages=113–118
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| |title=Characterizations of a Tangential Quadrilateral
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| |url=http://forumgeom.fau.edu/FG2009volume9/FG200910.pdf
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| |volume=9
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| |year=2009}}.</ref>
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| :<math>\tan{\frac{\angle ABD}{2}}\cdot\tan{\frac{\angle BDC}{2}}=\tan{\frac{\angle ADB}{2}}\cdot\tan{\frac{\angle DBC}{2}}.</math>
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| Further, a convex quadrilateral with successive sides ''a'', ''b'', ''c'', ''d'' is tangential if and only if
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| :<math>R_aR_c=R_bR_d</math>
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| where ''R''<sub>''a''</sub>, ''R''<sub>''b''</sub>, ''R''<sub>''c''</sub>, ''R''<sub>''d''</sub> are the radii in the circles externally tangent to the sides ''a'', ''b'', ''c'', ''d'' respectively and the extensions of the adjacent two sides for each side.<ref>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=63–77
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| |title=Similar Metric Characterizations of Tangential and Extangential Quadrilaterals
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| |url=http://forumgeom.fau.edu/FG2011volume12/FG201207.pdf
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| |volume=12
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| |year=2012}}.</ref>{{rp|p.72}}
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| Several [[Tangential quadrilateral#Characterizations in the four subtriangles|more characterizations]] are known in the four subtriangles formed by the diagonals.
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| ==Special line segments==
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| The eight ''tangent lengths'' of a tangential quadrilateral are the line segments from a [[vertex (geometry)|vertex]] to the points where the incircle is tangent to the sides. From each vertex there are two [[Congruence (geometry)|congruent]] tangent lengths.
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| The two ''tangency chords'' of a tangential quadrilateral are the line segments that connect points on opposite sides where the incircle is tangent to these sides. These are also the [[diagonal]]s of the ''contact quadrilateral''.
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| ==Area==
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| ===Non-trigonometric formulas===
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| The [[area]] ''K'' of a tangential quadrilateral is given by
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| :<math>\displaystyle K = r \cdot s,</math>
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| where ''s'' is the [[semiperimeter]] and ''r'' is the [[inradius]]. Another formula is<ref name=Durell/>
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| :<math>\displaystyle K = \tfrac{1}{2}\sqrt{p^2q^2-(ac-bd)^2}</math>
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| which gives the area in terms of the diagonals ''p'', ''q'' and the sides ''a'', ''b'', ''c'', ''d'' of the tangential quadrilateral.
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| The area can also be expressed in terms of just the four [[Tangential quadrilateral#Special line segments|tangent lengths]]. If these are ''e'', ''f'', ''g'', ''h'', then the tangential quadrilateral has the area<ref name=Josefsson/>
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| :<math>\displaystyle K=\sqrt{(e+f+g+h)(efg+fgh+ghe+hef)}.</math>
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| Furthermore, the area of a tangential quadrilateral can be expressed in terms of the sides ''a, b, c, d'' and the successive tangent lengths ''e, f, g, h'' as<ref name=Josefsson/>{{rp|p.128}}
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| :<math>K=\sqrt{abcd-(eg-fh)^2}.</math>
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| Since ''eg'' = ''fh'' if and only if the tangential quadrilateral is also cyclic and hence bicentric,<ref name=Hajja/> this shows that the maximal area <math>\sqrt{abcd}</math> occurs if and only if the tangential quadrilateral is bicentric.
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| ===Trigonometric formulas===
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| A [[trigonometry|trigonometric]] formula for the area in terms of the sides ''a'', ''b'', ''c'', ''d'' and two opposite angles is<ref name=Durell/><ref>Siddons, A.W., and R.T. Hughes, ''Trigonometry'', Cambridge Univ. Press, 1929: p. 203.</ref><ref name=Grinberg/><ref name=Yiu/>
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| :<math>\displaystyle K = \sqrt{abcd} \sin \frac{A+C}{2} = \sqrt{abcd} \sin \frac{B+D}{2}.</math>
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| For given side lengths, the area is [[Maxima and minima|maximum]] when the quadrilateral is also [[cyclic quadrilateral|cyclic]] and hence a [[bicentric quadrilateral]]. Then <math>K = \sqrt{abcd}</math> since opposite angles are [[supplementary angles]]. This can be proved in another way using [[calculus]].<ref>{{citation
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| |last=Hoyt |first=John P.
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| |journal=[[American Mathematical Monthly]]
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| |pages=54–56
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| |title=Maximizing the Area of a Trapezium
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| |volume=93
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| |number=1
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| |year=1986}}.</ref>
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| Another formula for the area of a tangential quadrilateral ''ABCD'' that involves two opposite angles is<ref name=Grinberg/>{{rp|p.19}}
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| :<math>K=\left(IA\cdot IC+IB\cdot ID\right)\sin\frac{A+C}{2}</math>
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| where ''I'' is the incenter.
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| In fact, the area can be expressed in terms of just two adjacent sides and two opposite angles as<ref name=Durell/>
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| :<math>K=ab\sin{\frac{B}{2}}\csc{\frac{D}{2}}\sin \frac{B+D}{2}.</math>
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| Still another area formula is<ref name=Durell>Durell, C.V. and Robson, A., ''Advanced Trigonometry'', Dover reprint, 2003, pp. 28–30.</ref>
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| :<math>K=\tfrac{1}{2}(ac-bd)\tan{\theta},</math>
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| where ''θ'' is the angle between the diagonals. This formula cannot be used when the tangential quadrilateral is a kite, since then ''θ'' is 90° and the tangent function is not defined.
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| ===Inequalities===
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| As indirectly noted above, the area of a tangential quadrilateral with sides ''a'', ''b'', ''c'', ''d'' satisfies
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| :<math>K\le\sqrt{abcd}</math>
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| with equality if and only if it is a [[bicentric quadrilateral]].
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| According to T. A. Ivanova (in 1976), the semiperimeter ''s'' of a tangential quadrilateral satisfies
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| :<math>s\ge 4r</math>
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| where ''r'' is the inradius. There is equality if and only if the quadrilateral is a [[Square (geometry)|square]].<ref>[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=466538 Post at ''Art of Problem Solving'', 2012]</ref> This means that for the area ''K'' = ''rs'', there is the [[inequality (mathematics)|inequality]]
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| :<math>K\ge 4r^2</math>
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| with equality if and only if the tangential quadrilateral is a square.
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| ===Partition properties===
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| [[File:Tangentenviereck.png|200px|thumb|Tangential quadrilateral with inradius ''r''.]]
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| The four line segments between the center of the incircle and the points where it is tangent to the quadrilateral partition the quadrilateral into four [[right kite]]s.
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| If a line cuts a tangential quadrilateral into two [[polygon]]s with equal [[area]]s and equal [[perimeter]]s, then that line passes through the [[incenter]].<ref name=Andreescu/>
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| ==Inradius==
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| The inradius in a tangential quadrilateral with consecutive sides ''a'', ''b'', ''c'', ''d'' is given by<ref name=Durell/>
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| :<math>r=\frac{K}{s}=\frac{K}{a+c}=\frac{K}{b+d}</math>
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| where ''K'' is the area of the quadrilateral and ''s'' is its semiperimeter. For a tangential quadrilateral with given sides, the inradius is [[Maxima and minima|maximum]] when the quadrilateral is also [[cyclic quadrilateral|cyclic]] (and hence a [[bicentric quadrilateral]]).
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| In terms of the [[Tangential quadrilateral#Special line segments|tangent lengths]], the incircle has radius<ref name=Hajja>{{citation
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| |last=Hajja |first=Mowaffaq
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| |journal=Forum Geometricorum
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| |pages=103–106
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| |title=A condition for a circumscriptible quadrilateral to be cyclic
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| |url=http://forumgeom.fau.edu/FG2008volume8/FG200814.pdf
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| |volume=8
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| |year=2008}}.</ref>{{rp|Lemma2}}<ref>{{citation
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| |last=Hoyt |first=John P.
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| |journal=[[Mathematics Magazine]]
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| |pages=239, 242
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| |title=Quickies, Q694
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| |volume=57
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| |number=4
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| |year=1984}}.</ref>
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| :<math>\displaystyle r=\sqrt{\frac{efg+fgh+ghe+hef}{e+f+g+h}}.</math>
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| The inradius can also be expressed in terms of the distances from the incenter ''I'' to the vertices of the tangential quadrilateral ''ABCD''. If ''u = AI'', ''v = BI'', ''x = CI'' and ''y = DI'', then
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| :<math>r=2\sqrt{\frac{(\sigma-uvx)(\sigma-vxy)(\sigma-xyu)(\sigma-yuv)}{uvxy(uv+xy)(ux+vy)(uy+vx)}}</math>
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| where <math>\sigma=\tfrac{1}{2}(uvx+vxy+xyu+yuv)</math>.<ref>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=27–34
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| |title=On the inradius of a tangential quadrilateral
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| |url=http://forumgeom.fau.edu/FG2010volume10/FG201005.pdf
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| |volume=10
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| |year=2010}}.</ref>
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| ==Angle formulas==
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| If ''e'', ''f'', ''g'' and ''h'' are the [[Tangential quadrilateral#Special line segments|tangent lengths]] from the vertices ''A'', ''B'', ''C'' and ''D'' respectively to the points where the incircle is tangent to the sides of a tangential quadrilateral ''ABCD'', then the [[angle]]s of the quadrilateral can be calculated from<ref name=Josefsson/>
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| :<math> \sin{\frac{A}{2}}=\sqrt{\frac{efg + fgh + ghe + hef}{(e + f)(e + g)(e + h)}},</math>
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| :<math> \sin{\frac{B}{2}}=\sqrt{\frac{efg + fgh + ghe + hef}{(f + e)(f + g)(f + h)}},</math>
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| :<math> \sin{\frac{C}{2}}=\sqrt{\frac{efg + fgh + ghe + hef}{(g + e)(g + f)(g + h)}},</math>
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| :<math> \sin{\frac{D}{2}}=\sqrt{\frac{efg + fgh + ghe + hef}{(h + e)(h + f)(h + g)}}.</math>
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| The angle between the [[Tangential quadrilateral#Special line segments|tangency chords]] ''k'' and ''l'' is given by<ref name=Josefsson/>
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| :<math> \sin{\varphi}=\sqrt{\frac{(e + f + g + h)(efg + fgh + ghe + hef)}{(e + f)(f + g)(g + h)(h + e)}}.</math>
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| ==Diagonals==
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| If ''e'', ''f'', ''g'' and ''h'' are the [[Tangential quadrilateral#Special line segments|tangent lengths]] from ''A'', ''B'', ''C'' and ''D'' respectively to the points where the incircle is tangent to the sides of a tangential quadrilateral ''ABCD'', then the lengths of the diagonals ''p = AC'' and ''q = BD'' are<ref name=Hajja/>{{rp|Lemma3}}
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| :<math>\displaystyle p=\sqrt{\frac{e+g}{f+h}\Big((e+g)(f+h)+4fh\Big)},</math>
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| :<math>\displaystyle q=\sqrt{\frac{f+h}{e+g}\Big((e+g)(f+h)+4eg\Big)}.</math>
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| ==Tangency chords==
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| If ''e'', ''f'', ''g'' and ''h'' are the [[Tangential quadrilateral#Special line segments|tangent lengths]] of a tangential quadrilateral, then the lengths of the [[Tangential quadrilateral#Special line segments|tangency chords]] are<ref name=Josefsson/>
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| :<math>\displaystyle k=\frac{2(efg+fgh+ghe+hef)}{\sqrt{(e+f)(g+h)(e+g)(f+h)}},</math>
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| :<math>\displaystyle l=\frac{2(efg+fgh+ghe+hef)}{\sqrt{(e+h)(f+g)(e+g)(f+h)}}</math>
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| where the tangency chord of length ''k'' connects the sides of lengths ''a'' = ''e'' + ''f'' and ''c'' = ''g'' + ''h'', and the one of length ''l'' connects the sides of lengths ''b'' = ''f'' + ''g'' and ''d'' = ''h'' + ''e''. The squared ratio of the tangency chords satisfy<ref name=Josefsson/>
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| :<math>\frac{k^2}{l^2} = \frac{bd}{ac}.</math>
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| The two tangency chords
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| *are [[perpendicular]] if and only if the tangential quadrilateral also has a [[circumcircle]] (it is [[Bicentric quadrilateral|bicentric]]).<ref name=Josefsson/>{{rp|p.124}}
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| *have equal lengths if and only if the tangential quadrilateral is a [[Kite (geometry)|kite]].<ref name=Josefsson3/>{{rp|p.166}}
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| The tangency chord between the sides ''AB'' and ''CD'' in a tangential quadrilateral ''ABCD'' is longer than the one between the sides ''BC'' and ''DA'' if and only if the [[Quadrilateral#Special line segments|bimedian]] between the sides ''AB'' and ''CD'' is shorter than the one between the sides ''BC'' and ''DA''.<ref>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=155–164
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| |title=The Area of a Bicentric Quadrilateral
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| |url=http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf
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| |volume=11
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| |year=2011}}.</ref>{{rp|p.162}}
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| If tangential quadrilateral ''ABCD'' has tangency points ''W'' on ''AB'' and ''Y'' on ''CD'', and if tangency chord ''WY'' intersects diagonal ''BD'' at ''M'', then the ratio of tangent lengths <math>\tfrac{BW}{DY}</math> equals the ratio <math>\tfrac{BM}{DM}</math> of the segments of diagonal ''BD''.<ref>Gutierrez, Antonio, "Circumscribed Quadrilateral, Diagonal, Chord, Proportion", [http://gogeometry.com/problem/p152_circumscribed_quadrilateral_diagonal_chord.htm], Accessed 2012-04-09.</ref>
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| ==Collinear points==
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| If ''M<sub>1</sub>'' and ''M<sub>2</sub>'' are the [[midpoint]]s of the diagonals ''AC'' and ''BD'' respectively in a tangential quadrilateral ''ABCD'' with incenter ''I'', and if the pairs of opposite sides meet at ''J'' and ''K'' with ''M<sub>3</sub>'' being the midpoint of ''JK'', then the points ''M<sub>3</sub>'', ''M<sub>1</sub>'', ''I'', and ''M<sub>2</sub>'' are [[Collinearity|collinear]].<ref name=Andreescu/>{{rp|p.42}} The line containing them is the [[Newton line]] of the quadrilateral.
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| If the extensions of opposite sides in a tangential quadrilateral intersect at ''J'' and ''K'', and the extensions of opposite sides in its contact quadrilateral intersect at ''L'' and ''M'', then the four points ''J'', ''L'', ''K'' and ''M'' are collinear.<ref name=Josefsson4>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=165–173
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| |title=Characterizations of Bicentric Quadrilaterals
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| |url=http://forumgeom.fau.edu/FG2010volume10/FG201019.pdf
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| |volume=10
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| |year=2010}}.</ref>{{rp|Cor.3}}
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| If the incircle is tangent to the sides ''AB'', ''BC'', ''CD'', ''DA'' at ''T<sub>1</sub>'', ''T<sub>2</sub>'', ''T<sub>3</sub>'', ''T<sub>4</sub>'' respectively, and if ''N<sub>1</sub>'', ''N<sub>2</sub>'', ''N<sub>3</sub>'', ''N<sub>4</sub>'' are the [[isotomic conjugate]]s of these points with respect to the corresponding sides (that is, ''AT<sub>1</sub>'' = ''BN<sub>1</sub>'' and so on), then the ''Nagel point'' of the tangential quadrilateral is defined as the intersection of the lines ''N<sub>1</sub>N<sub>3</sub>'' and ''N<sub>2</sub>N<sub>4</sub>''. Both of these lines divide the [[perimeter]] of the quadrilateral into two equal parts. More importantly, the Nagel point ''N'', the [[Quadrilateral#Remarkable points and lines in a convex quadrilateral|"area centroid"]] ''G'', and the incenter ''I'' are collinear in this order, and ''NG'' = 2''GI''. This line is called the ''Nagel line'' of a tangential quadrilateral.<ref name=Myakishev>{{citation
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| | last = Myakishev | first = Alexei
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| | journal = Forum Geometricorum
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| | pages = 289–295
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| | title = On Two Remarkable Lines Related to a Quadrilateral
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| | url = http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf
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| | volume = 6
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| | year = 2006}}.</ref>
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| In a tangential quadrilateral ''ABCD'' with incenter ''I'' and where the diagonals intersect at ''P'', let ''H<sub>X</sub>'', ''H<sub>Y</sub>'', ''H<sub>Z</sub>'', ''H<sub>W</sub>'' be the [[Altitude (triangle)|orthocenter]]s of triangles ''AIB'', ''BIC'', ''CID'', ''DIA''. Then the points ''P'', ''H<sub>X</sub>'', ''H<sub>Y</sub>'', ''H<sub>Z</sub>'', ''H<sub>W</sub>'' are collinear.<ref name=Grinberg/>{{rp|p.28}}
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| ==Concurrent and perpendicular lines==
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| The two diagonals and the two tangency chords are [[Concurrent lines|concurrent]].<ref name=Yiu>Yiu, Paul, ''Euclidean Geometry'', [http://www.math.fau.edu/Yiu/EuclideanGeomeryNotes.pdf], 1998, pp. 156–157.</ref> One way to see this is as a limiting case of [[Brianchon's theorem]], which states that a hexagon all of whose sides are tangent to a single [[conic section]] has three diagonals that meet at a point. From a tangential quadrilateral, one can form a hexagon with two 180° angles, by placing two new vertices at two opposite points of tangency; all six of the sides of this hexagon lie on lines tangent to the inscribed circle, so its diagonals meet at a point. But two of these diagonals are the same as the diagonals of the tangential quadrilateral, and the third diagonal of the hexagon is the line through two opposite points of tangency. Repeating this same argument with the other two points of tangency completes the proof of the result.
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| If the incircle is tangent to the sides ''AB'', ''BC'', ''CD'', ''DA'' at ''W'', ''X'', ''Y'', ''Z'' respectively, then the lines ''WX'', ''ZY'' and ''AC'' are concurrent.<ref name=Grinberg>[http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf Grinberg, Darij, ''Circumscribed quadrilaterals revisited'', 2008]</ref>{{rp|p.11}}
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| If the extensions of opposite sides in a tangential quadrilateral intersect at ''J'' and ''K'', and the diagonals intersect at ''P'', then ''JK'' is perpendicular to the extension of ''IP'' where ''I'' is the incenter.<ref name=Josefsson4/>{{rp|Cor.4}}
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| ==Metric properties of the incenter==
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| The ratio of two opposite sides in a tangential quadrilateral can be expressed in terms of the distances between the incenter ''I'' and the vertices according to<ref name=Grinberg/>{{rp|p.15}}
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| :<math>\frac{AB}{CD}=\frac{IA\cdot IB}{IC\cdot ID},\quad\quad \frac{BC}{DA}=\frac{IB\cdot IC}{ID\cdot IA}.</math>
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| The product of two adjacent sides in a tangential quadrilateral ''ABCD'' with incenter ''I'' satisfies<ref>"Ineq-G126 - Geometry - very nice!!!!", Post at ''Art of Problem Solving'', 2011, [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=403668]</ref>
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| :<math>AB\cdot BC=IB^2+\frac{IA\cdot IB\cdot IC}{ID}.</math>
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| If ''I'' is the incenter of a tangential quadrilateral ''ABCD'', then<ref name=Grinberg/>{{rp|p.16}}
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| :<math>IA\cdot IC+IB\cdot ID=\sqrt{AB\cdot BC\cdot CD\cdot DA}.</math>
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| The incenter ''I'' in a tangential quadrilateral ''ABCD'' coincides with the [[Quadrilateral#Remarkable points and lines in a convex quadrilateral|"vertex centroid"]] of the quadrilateral [[if and only if]]<ref name=Grinberg/>{{rp|p.22}}
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| :<math>IA\cdot IC=IB\cdot ID.</math>
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| If ''M<sub>p</sub>'' and ''M<sub>q</sub>'' are the [[midpoint]]s of the diagonals ''AC'' and ''BD'' respectively in a tangential quadrilateral ''ABCD'' with incenter ''I'', then <ref name=Grinberg/>{{rp|p.19}}<ref>[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=455293 "Determine ratio OM/ON", Post at ''Art of Problem Solving'', 2011]</ref>
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| :<math>\frac{IM_p}{IM_q}=\frac{IA\cdot IC}{IB\cdot ID}=\frac{e+g}{f+h}</math>
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| where ''e'', ''f'', ''g'' and ''h'' are the tangent lengths at ''A'', ''B'', ''C'' and ''D'' respectively. Combining the first equality with a previous property, the "vertex centroid" of the tangential quadrilateral coincides with the incenter if and only if the incenter is the midpoint of the line segment connecting the midpoints of the diagonals. | |
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| If a [[four-bar linkage]] is made in the form of a tangential quadrilateral, then it will remain tangential no matter how the linkage is flexed, provided the quadrilateral remains convex.<ref>{{citation |first=Helen |last=Barton |title=On a circle attached to a collapsible four-bar |journal=[[American Mathematical Monthly]] |volume=33 |issue=9 |year=1926 |pages=462–465 |jstor=2299611}}.</ref><ref>Bogomolny, Alexander, "When A Quadrilateral Is Inscriptible?", ''Interactive Mathematics Miscellany and Puzzles'', [http://www.cut-the-knot.org/proofs/InscriptibleQuadrilateral.shtml#anotherProof].</ref> (Thus, for example, if a square is deformed into a rhombus it remains tangential, though to a smaller incircle). If one side is held in a fixed position, then as the quadrilateral is flexed, the incenter traces out a circle of radius <math>\sqrt{abcd}/s</math> where ''a,b,c,d'' are the sides in sequence and ''s'' is the semiperimeter.
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| ==Characterizations in the four subtriangles==
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| [[Image:Chao tangentual quad radii.svg|thumb|240px|Chao and Simeonov's characterization in terms of the radii of circles within each of four triangles]]
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| In the nonoverlapping triangles ''APB'', ''BPC'', ''CPD'', ''DPA'' formed by the diagonals in a convex quadrilateral ''ABCD'', where the diagonals intersect at ''P'', there are the following characterizations of tangential quadrilaterals.
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| Let ''r''<sub>1</sub>, ''r''<sub>2</sub>, ''r''<sub>3</sub>, and ''r''<sub>4</sub> denote the radii of the incircles in the four triangles ''APB'', ''BPC'', ''CPD'', and ''DPA'' respectively. Chao and Simeonov proved that the quadrilateral is tangential [[if and only if]]<ref>{{citation
| |
| |last1=Chao |first1=Wu Wei |last2=Simeonov |first2=Plamen
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| |title=When quadrilaterals have inscribed circles (solution to problem 10698)
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| |journal=[[American Mathematical Monthly]]
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| |year=2000
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| |volume=107
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| |issue=7
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| |pages=657–658
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| |doi=10.2307/2589133}}.</ref>
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| :<math>\frac{1}{r_1}+\frac{1}{r_3}=\frac{1}{r_2}+\frac{1}{r_4}.</math>
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| This characterization had already been proved five years earlier by Vaynshtejn.<ref name=Josefsson3/>{{rp|p.169}}<ref name=Vaynshtejn>{{citation
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| |last1=Vaynshtejn |first1=I. |last2=Vasilyev |first2=N. |last3=Senderov |first3=V.
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| |title=(Solution to problem) M1495
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| |journal=Kvant
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| |year=1995
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| |issue=6
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| |pages=27–28.}}</ref>
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| In the solution to his problem, a similar characterization was given by Vasilyev and Senderov. If ''h''<sub>1</sub>, ''h''<sub>2</sub>, ''h''<sub>3</sub>, and ''h''<sub>4</sub> denote the [[altitude (triangle)|altitude]]s in the same four triangles (from the diagonal intersection to the sides of the quadrilateral), then the quadrilateral is tangential if and only if<ref name=Minculete/><ref name=Vaynshtejn/>
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| :<math>\frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}.</math> | |
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| Another similar characterization concerns the [[Incircle and excircles of a triangle#Relation to area of the triangle|exradii]] ''r''<sub>''a''</sub>, ''r''<sub>''b''</sub>, ''r''<sub>''c''</sub>, and ''r''<sub>''d''</sub> in the same four triangles (the four [[Incircle and excircles of a triangle|excircle]]s are each tangent to one side of the quadrilateral and the extensions of its diagonals). A quadrilateral is tangential if and only if<ref name=Josefsson2/>{{rp|p.70}}
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| :<math>\frac{1}{r_a}+\frac{1}{r_c}=\frac{1}{r_b}+\frac{1}{r_d}.</math>
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| If ''R''<sub>1</sub>, ''R''<sub>2</sub>, ''R''<sub>3</sub>, and ''R''<sub>4</sub> denote the radii in the [[circumcircle]]s of triangles ''APB'', ''BPC'', ''CPD'', and ''DPA'' respectively, then the quadrilateral ''ABCD'' is tangential if and only if<ref>{{citation
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| | last = Josefsson | first = Martin
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| | journal = Forum Geometricorum
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| | pages = 13–25
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| | title = Characterizations of Orthodiagonal Quadrilaterals
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| | url = http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf
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| | volume = 12
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| | year = 2012}}.</ref>{{rp|pp. 23–24}}
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| :<math>R_1+R_3=R_2+R_4.</math>
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| In 1996, Vaynshtejn was probably the first to prove another beautiful characterization of tangential quadrilaterals, that has later appeared in several magazines and websites.<ref name=Josefsson2/>{{rp|pp. 72–73}} It states that when a convex quadrilateral is divided into four nonoverlapping triangles by its two diagonals, then the incenters of the four triangles are concyclic if and only if the quadrilateral is tangential. In fact, the incenters form an [[Cyclic quadrilateral#Properties of cyclic quadrilaterals that are also orthodiagonal|orthodiagonal cyclic quadrilateral]].<ref name=Josefsson2/>{{rp|p.74}} A related result is that the incircles can be exchanged for the excircles to the same triangles (tangent to the sides of the quadrilateral and the extensions of its diagonals). Thus a convex quadrilateral is tangential if and only if the excenters in these four [[Incircle and excircles of a triangle|excircle]]s are the vertices of a [[cyclic quadrilateral]].<ref name=Josefsson2/>{{rp|p. 73}}
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| A convex quadrilateral ''ABCD'', with diagonals intersecting at ''P'', is tangential if and only if the four excenters in triangles ''APB'', ''BPC'', ''CPD'', and ''DPA'' opposite the vertices ''B'' and ''D'' are concyclic.<ref name=Josefsson2/>{{rp|p. 79}} If ''R<sub>a</sub>'', ''R<sub>b</sub>'', ''R<sub>c</sub>'', and ''R<sub>d</sub>'' are the exradii in the triangles ''APB'', ''BPC'', ''CPD'', and ''DPA'' respectively opposite the vertices ''B'' and ''D'', then another condition is that the quadrilateral is tangential if and only if<ref name=Josefsson2/>{{rp|p. 80}}
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| :<math>\frac{1}{R_a}+\frac{1}{R_c}=\frac{1}{R_b}+\frac{1}{R_d}.</math>
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| Further, a convex quadrilateral ''ABCD'' with diagonals intersecting at ''P'' is tangential if and only if<ref name=Minculete/>
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| :<math>\frac{a}{\text{∆}(APB)}+\frac{c}{\text{∆}(CPD)}=\frac{b}{\text{∆}(BPC)}+\frac{d}{\text{∆}(DPA)}</math>
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| where ∆(''APB'') is the area of triangle ''APB''.
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| Denote the segments that the diagonal intersection ''P'' divides diagonal ''AC'' into as ''AP'' = ''p''<sub>1</sub> and ''PC'' = ''p''<sub>2</sub>, and similarly ''P'' divides diagonal ''BD'' into segments ''BP'' = ''q''<sub>1</sub> and ''PD'' = ''q''<sub>2</sub>. Then the quadrilateral is tangential if and only if any one of the following equalities are true:<ref>{{citation
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| | last = Hoehn | first = Larry
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| | journal = Forum Geometricorum
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| | pages = 211–212
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| | title = A new formula concerning the diagonals and sides of a quadrilateral
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| | url = http://forumgeom.fau.edu/FG2011volume11/FG201122.pdf
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| | volume = 11
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| | year = 2011}}.</ref>
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| :<math>ap_2q_2 + cp_1q_1 = bp_1q_2 + dp_2q_1</math>
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| or<ref name=Josefsson2/>{{rp|p. 74}}
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| :<math>\frac{(p_1+q_1-a)(p_2+q_2-c)}{(p_1+q_1+a)(p_2+q_2+c)}=\frac{(p_2+q_1-b)(p_1+q_2-d)}{(p_2+q_1+b)(p_1+q_2+d)}</math>
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| or<ref name=Josefsson2/>{{rp|p. 77}}
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| :<math>\frac{(a+p_1-q_1)(c+p_2-q_2)}{(a-p_1+q_1)(c-p_2+q_2)}=\frac{(b+p_2-q_1)(d+p_1-q_2)}{(b-p_2+q_1)(d-p_1+q_2)}.</math>
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| ==Conditions for a tangential quadrilateral to be another type of quadrilateral==
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| A tangential quadrilateral is a [[rhombus]] if and only if its opposite angles are equal.<ref>De Villiers, Michael, Equiangular cyclic and equilateral circumscribed polygons", ''[[Mathematical Gazette]]'' 95, March 2011, 102–107.</ref>
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| If the incircle is tangent to the sides ''AB'', ''BC'', ''CD'', ''DA'' at ''W'', ''X'', ''Y'', ''Z'' respectively, then a tangential quadrilateral ''ABCD'' is also [[Cyclic quadrilateral|cyclic]] (and hence [[bicentric quadrilateral|bicentric]]) if and only if any of<ref name=Bryant/><ref name=Josefsson/>{{rp|p.124}}<ref name=Josefsson4/>
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| *''WY'' is perpendicular to ''XZ''
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| *<math>\frac{AW}{WB}=\frac{DY}{YC}</math>
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| *<math>\frac{AC}{BD}=\frac{AW+CY}{BX+DZ}</math>
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| The first of these three means that the ''contact quadrilateral'' ''WXYZ'' is an [[orthodiagonal quadrilateral]].
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| A tangential quadrilateral is a [[Kite (geometry)|kite]] if and only if any one of the following conditions is true:<ref name=Josefsson3>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=165–174
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| |title=When is a Tangential Quadrilateral a Kite?
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| |url=http://forumgeom.fau.edu/FG2011volume11/FG201117.pdf
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| |volume=11
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| |year=2011}}.</ref>
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| *The area is one half the product of the [[diagonal]]s
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| *The diagonals are [[perpendicular]]
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| *The two line segments connecting opposite points of tangency have equal length
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| *One pair of opposite [[Tangential quadrilateral#Special_line_segments|tangent lengths]] have equal length
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| *The [[Quadrilateral#Special line segments|bimedians]] have equal length
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| *The products of opposite sides are equal
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| *The center of the incircle lies on the diagonal that is the axis of symmetry.
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| ==See also==
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| *[[Circumscribed circle]]
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| *[[Ex-tangential quadrilateral]]
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| *[[Tangential trapezoid]]
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| ==References==
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| {{reflist}}
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| ==External links==
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| *{{MathWorld |title=Tangential Quadrilateral |urlname=TangentialQuadrilateral}}
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| {{DEFAULTSORT:Tangential Quadrilateral}}
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| [[Category:Quadrilaterals]]
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