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| :'' '''Incircle''' redirects here. For incircles of non-triangle polygons, see [[Tangential quadrilateral]] or [[Tangential polygon]].
| | Although not especially African American skin care products, the formulas that contain the ingredients I have listed here will work extremely well in alleviating the common problems you suffer. Skin cleansers, moisturizers, toners, and facial masks. , people eat health lack of due Vitiligo, abdomen, supplements, steel along with supplements Vitiligo going on a fast mortgage. These autoimmune diseases include hyperthyroidism (an overactive thyroid gland), adrenocortical insufficiency (the adrenal gland does not produce enough of the hormone called corticosteroid), alopecia areata (patches of baldness), and pernicious anemia (a low level of red blood cells caused by failure of the body to absorb vitamin B-12). One of the things that there is a high incidence of is acne, which is caused in part by excessive oil production. <br><br>The cause of vitiligo is not known, researchers have numerous different theories. These involve corticosteroid lotions or the topical immunomodulator, tacrolimus. Vitiligo is a skin disorder that triggers white spots of epidermis to appear on different components of your body. Powder around 35 gm of the seeds of radish and mix it in 2 tsp of vinegar. In some cases, especially when it occurs suddenly and aggressively, hirsutism might signalize cancer or a tumor. <br><br>For complete vitiligo cure, this is an excellent herbal product. Vitiligo is a Latina language word whose indicating is calf. The major symptoms & signs of the disease are appearance of white patches on different skin areas. Gray or white hair can be restored, reversed, and remedied with a topical application known as a "proprietary treatment. Otherwise, the condition is completely asmptomatis with no structural change or loss of sensation in the affected areas. <br><br>In early stage it can be appeared in some particular area of skin, small in size and after that it can be spread on whole body. In case of a few patients, symptoms get visible in very few days whereas in some cases you cant recognize it in early stage. You can also have it directly 2 times daily, once in the morning and again in the evening for 2 months continuously. It is a skin disease in which some areas of our skin become white due to the gradual depigmentation or loss of melanin from skin layer. It is very important fight one other issues that Vitiligo Treatment brings alongside. <br><br>There are plenty of otions for treatment for leucoderma. Advise patients to use corrective cosmetics to help hide skin lesions, and to use a sunblock because hypopigmented areas may sunburn easily. PUVA (Phototherapy Ultra Violet A) is the most remarkable instance in which ultraviolet radiation play a crucial role along with oral or local Psoralen therapeutic modes. Always wear sunscreen every day, and as much as possible to stay home and avoid exposure to sunlight. Coconut oil is one of the most commonly used hair treatment. <br><br>It is a skin disorder mainly due to inappropriate diet combination intake regularly and formed due to disturbance of tridosas namely 'Vaat', 'Pitta' and 'Kapha'. Also, it is best to use physical sunblock only, or sunblock that sits atop the skin and only use a recommended moisturizer. ' Mixed Vitiligo: If more than one patterns of distribution are establish in the same individual, like segmental and acrofacial, it is recognized as Mixed Vitiligo. Associated effects may include very painful, palpable peripheral nerves; muscle atrophy and contractures; and ulcers of the fingers and toes. Vitiligo word is derived from Latin word "VITIUM" (blemish) and the suffix- "IGO".<br><br>If you have any queries pertaining to exactly where and how to use [http://discover-prague.info/ vitiligo home treatment], you can get in touch with us at our web-site. |
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| [[Image:Incircle and Excircles.svg|right|thumb|300px|A triangle (black) with incircle (blue), [[incenter]] (I), excircles (orange), excenters (J<sub>A</sub>,J<sub>B</sub>,J<sub>C</sub>), internal [[angle bisector]]s (red) and external angle bisectors (green)]]
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| In [[geometry]], the '''incircle''' or '''inscribed circle''' of a [[triangle]] is the largest [[circle]] contained in the triangle; it touches (is [[tangent]] to) the three sides. The center of the incircle is called the triangle's '''incenter'''.
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| An '''excircle''' or '''escribed circle''' of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two.
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| Every triangle has three distinct excircles, each tangent to one of the triangle's sides.
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| The center of the incircle can be found as the intersection of the three internal [[angle bisector]]s.
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| The center of an excircle is the intersection of the internal bisector of one angle and the external bisectors of the other two. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an [[orthocentric system]].
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| See also [[Tangent lines to circles]].
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| == Relation to area of the triangle ==
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| The radii of the in- and excircles are closely related to the [[area]] of the triangle. Let ''K'' be the triangle's area and let ''a'', ''b'' and ''c'', be the lengths of its sides. By [[Heron's formula]], the area of the triangle is
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| :<math>
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| \begin{align}
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| K & {} = \frac{1}{4}\sqrt{(P)(a-b+c)(b-c+a)(c-a+b)} \\
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| & {} = \sqrt{s(s-a)(s-b)(s-c)}
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| \end{align}
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| </math>
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| where <math>s= \tfrac{1}{2}(a+b+c)</math> is the semiperimeter and ''P'' = 2''s'' is the perimeter.
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| The radius of the incircle (also known as the '''inradius''', ''r'' ) is
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| : <math>r = \frac{2K}{P} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math>
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| Thus, the area ''K'' of a triangle may be found by multiplying the inradius by the semiperimeter:
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| :<math>\displaystyle K=rs.</math>
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| The radii in the excircles are called the '''exradii'''. The excircle at side ''a'' has radius
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| : <math>r_a = \frac{2K}{c-a+b} = \sqrt{\frac{s (s-b)(s-c)}{s-a}}.</math>
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| Similarly the radii of the excircles at sides ''b'' and ''c'' are respectively
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| : <math>r_b = \frac{2K}{a-b+c} = \sqrt{\frac{s (s-a)(s-c)}{s-b}}</math>
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| and | |
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| : <math>r_c = \frac{2K}{b-c+a} = \sqrt{\frac{s (s-a)(s-b)}{s-c}}.</math>
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| From these formulas one can see that the excircles are always larger than the incircle and that the largest excircle is the one tangent to the longest side and the smallest excircle is tangent to the shortest side. Further, combining these formulas with Heron's area formula yields the result that<ref>Baker, Marcus, "A collection of formulae for the area of a plane triangle," ''Annals of Mathematics'', part 1 in vol. 1(6), January 1885, 134-138. (See also part 2 in vol. 2(1), September 1885, 11-18.)</ref>{{rp|#4}}
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| :<math>K=\sqrt{rr_ar_br_c}.</math>
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| The ratio of the area of the incircle to the area of the triangle is less than or equal to <math>\frac{\pi}{3\sqrt{3}}</math>, with equality holding only for [[equilateral triangle]]s.<ref>Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", ''[[American Mathematical Monthly]]'' 115, October 2008, 679-689: Theorem 4.1.</ref>
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| == Nine-point circle and Feuerbach point ==
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| The circle tangent to all three of the excircles as well as the incircle is known as the [[nine-point circle]]. The point where the nine-point circle touches the incircle is known as the Feuerbach point.
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| == Gergonne triangle and point ==
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| [[Image:Intouch Triangle and Gergonne Point.svg|right|frame|200px|A triangle, Δ''ABC'', with incircle (blue), incenter (blue, ''I''), contact triangle (red, Δ''T''<sub>''a''</sub>''T''<sub>''b''</sub>''T''<sub>''c''</sub>) and Gergonne point (green, Ge)]]
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| The '''Gergonne triangle'''(of ''ABC'') is defined by the 3 touchpoints of the incircle on the 3 sides.
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| Those vertices are denoted as ''T<sub>A</sub>'', etc.
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| The point that ''T<sub>A</sub>'' denotes, lies opposite to ''A''.
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| This '''Gergonne triangle''' ''T<sub>A</sub>T<sub>B</sub>T<sub>C</sub>'' is also known as the '''contact triangle''' or '''intouch triangle''' of ''ABC''.
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| The three lines ''AT<sub>A</sub>'', ''BT<sub>B</sub>'' and ''CT<sub>C</sub>'' intersect in a single point called '''Gergonne point''', denoted as ''Ge'' - [[Triangle center|''X(7)'']].
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| Interestingly, the Gergonne point of a triangle is the [[symmedian point]] of the Gergonne triangle. For a full set of properties of the Gergonne point see.<ref>{{Cite journal
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| | last = Dekov
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| | first = Deko
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| | title = Computer-generated Mathematics : The Gergonne Point
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| | journal = Journal of Computer-generated Euclidean Geometry
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| | year = 2009
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| | volume = 1
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| | pages = 1–14.
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| | url = http://www.dekovsoft.com/j/2009/01/JCGEG200901.pdf}}</ref>
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| The touchpoints of the three excircles with segments ''BC,CA and AB'' are the vertices of the [[extouch triangle]]. The points of intersection of the interior angle bisectors of ''ABC'' with the segments ''BC,CA,AB'' are the vertices of the '''incentral triangle'''. | |
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| == Nagel triangle and point ==
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| The '''Nagel triangle''' of ''ABC'' is denoted by the vertices ''X<sub>A</sub>'', ''X<sub>B</sub>'' and ''X<sub>C</sub>'' that are the three points where the excircles touch the reference triangle ''ABC'' and where ''X<sub>A</sub>'' is opposite of ''A'', etc. This triangle ''X<sub>A</sub>X<sub>B</sub>X<sub>C</sub>'' is also known as the '''extouch triangle''' of ''ABC''. The circumcircle of the extouch triangle ''X<sub>A</sub>X<sub>B</sub>X<sub>C</sub>'' is called the '''Mandart circle'''. The three lines ''AX<sub>A</sub>'', ''BX<sub>B</sub>'' and ''CX<sub>C</sub>'' are called the [[Splitter (geometry)|splitters]] of the triangle; they each bisect the perimeter of the triangle, and they intersect in a single point, the triangle's [[Nagel point]] ''Na'' - [[Triangle center|''X(8)'']].
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| [[Trilinear coordinates]] for the vertices of the intouch triangle are given by
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| *<math> A-\text{vertex}= 0 : \sec^2 \left(\frac{B}{2}\right) :\sec^2\left(\frac{C}{2}\right)</math>
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| *<math> B-\text{vertex}= \sec^2 \left(\frac{A}{2}\right):0:\sec^2\left(\frac{C}{2}\right)</math>
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| *<math> C-\text{vertex}= \sec^2 \left(\frac{A}{2}\right) :\sec^2\left(\frac{B}{2}\right):0</math>
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| Trilinear coordinates for the vertices of the extouch triangle are given by
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| * <math> A-\text{vertex} = 0 : \csc^2\left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)</math>
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| * <math> B-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : 0 : \csc^2\left(\frac{C}{2}\right)</math>
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| * <math> C-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : \csc^2\left(\frac{B}{2}\right) : 0</math>
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| Trilinear coordinates for the vertices of the incentral triangle are given by
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| * <math>\ A-\text{vertex} = 0 : 1 : 1</math>
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| * <math>\ B-\text{vertex} = 1 : 0 : 1</math>
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| * <math>\ C-\text{vertex} = 1 : 1 : 0</math>
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| Trilinear coordinates for the vertices of the excentral triangle are given by
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| * <math>\ A-\text{vertex}= -1 : 1 : 1 </math>
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| * <math>\ B-\text{vertex}= 1 : -1 : 1 </math>
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| * <math>\ C-\text{vertex}= 1 : -1 : -1 </math>
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| Trilinear coordinates for the Gergonne point are given by
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| : <math>\sec^2\left(\frac{A}{2}\right) : \sec^2 \left(\frac{B}{2}\right) : \sec^2\left(\frac{C}{2}\right)</math>,
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| or, equivalently, by the [[Law of Sines]], | |
| : <math>\frac{bc}{b+ c - a} : \frac{ca}{c + a-b} : \frac{ab}{a+b-c}</math>.
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| Trilinear coordinates for the Nagel point are given by
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| : <math>\csc^2\left(\frac{A}{2}\right) : \csc^2 \left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)</math>,
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| or, equivalently, by the [[Law of Sines]],
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| : <math>\frac{b+ c - a}{a} : \frac{c + a-b}{b} : \frac{a+b-c}{c}</math>.
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| It is the isotomic conjugate of the Gergonne point.
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| ==Coordinates of the incenter==
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| The [[Cartesian coordinates]] of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle as weights. (The weights are positive so the incenter lies inside the triangle as stated above.) If the three vertices are located at <math>(x_a,y_a)</math>, <math>(x_b,y_b)</math>, and <math>(x_c,y_c)</math>, and the sides opposite these vertices have corresponding lengths <math>a</math>, <math>b</math>, and <math>c</math>, then the incenter is at | |
| :<math>\bigg(\frac{a x_a+b x_b+c x_c}{P},\frac{a y_a+b y_b+c y_c}{P}\bigg) = \frac{a(x_a,y_a)+b(x_b,y_b)+c(x_c,y_c)}{P}</math>
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| where <math>\ P = a + b + c.</math>
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| [[Trilinear coordinates]] for the incenter are given by
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| :<math>\ 1 : 1 : 1.</math>
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| [[Barycentric coordinates (mathematics)|Barycentric coordinates]] for the incenter are given by
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| :<math>\ a : b : c</math>
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| or equivalently
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| :<math>\sin(A):\sin(B):\sin(C).</math>
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| ==Equations for four circles==
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| Let x : y : z be a variable point in [[trilinear coordinates]], and let u = cos<sup>''2''</sup>''(A/2)'', v = cos<sup>''2''</sup>''(B/2)'', w = cos<sup>''2''</sup>''(C/2)''. The four circles described above are given by these equations:
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| :* Incircle:
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| ::<math>\ u^2x^2+v^2y^2+w^2z^2-2vwyz-2wuzx-2uvxy=0</math>
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| :* ''A-''excircle:
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| ::<math>\ u^2x^2+v^2y^2+w^2z^2-2vwyz+2wuzx+2uvxy=0</math>
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| :* ''B-''excircle:
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| ::<math>\ u^2x^2+v^2y^2+w^2z^2+2vwyz-2wuzx+2uvxy=0</math>
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| :* ''C-''excircle:
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| ::<math>\ u^2x^2+v^2y^2+w^2z^2+2vwyz+2wuzx-2uvxy=0</math>
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| ==Euler's theorem==
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| [[Euler's theorem in geometry|Euler's theorem]] states that in a triangle:
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| :<math>(R-r_{in})^2=d^2+r_{in}^2,</math>
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| where ''R'' and ''r''<sub>''in''</sub> are the circumradius and inradius respectively, and ''d'' is the distance between the circumcenter and the incenter.
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| For excircles the equation is similar:
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| :<math>(R+r_{ex})^2=d^2+r_{ex}^2,</math>
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| where ''r''<sub>''ex''</sub> is the radius one of the excircles, and ''d'' is the distance between the circumcenter and this excircle's center.
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| <ref name=Nelson>Nelson, Roger, "Euler's triangle inequality via proof without words," ''Mathematics Magazine'' 81(1), February 2008, 58-61.</ref> | |
| <ref>Johnson, R. A. ''Modern Geometry'', Houghton Mifflin, Boston, 1929: p. 187.</ref> | |
| <ref>[http://forumgeom.fau.edu/FG2001volume1/FG200120.pdf Emelyanov, Lev, and Emelyanova, Tatiana. "Euler’s formula and Poncelet’s porism", ''Forum Geometricorum'' 1, 2001: pp. 137–140.]</ref>
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| ==Other incircle properties==
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| Suppose the tangency points of the incircle divide the sides into lengths of ''x'' and ''y'', ''y'' and ''z'', and ''z'' and ''x''. Then the incircle has the radius<ref>Chu, Thomas, ''The Pentagon'', Spring 2005, p. 45, problem 584.</ref>
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| :<math> r = \sqrt{\frac{xyz}{x+y+z}}</math>
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| and the area of the triangle is
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| :<math>K=\sqrt{xyz(x+y+z)}.</math>
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| If the altitudes from sides of lengths ''a'', ''b'', and ''c'' are ''h<sub>a</sub>'', ''h<sub>b</sub>'', and ''h<sub>c</sub>'' then the inradius ''r'' is one-third of the harmonic mean of these altitudes, i.e.
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| :<math> r = \frac{1}{h_a^{-1}+h_b^{-1}+h_c^{-1}}.</math>
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| The product of the incircle radius ''r'' and the circumcircle radius ''R'' of a triangle with sides ''a'', ''b'', and ''c'' is<ref>Johnson, Roger A., ''Advanced Euclidean Geometry'', Dover, 2007 (orig. 1929), p. 189, #298(d).</ref>
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| :<math>rR=\frac{abc}{2(a+b+c)}.</math>
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| Some relations among the sides, incircle radius, and circumcircle radius are:<ref name=Bell/>
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| :<math>ab+bc+ca=s^2+(4R+r)r,</math>
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| :<math>a^2+b^2+c^2=2s^2-2(4R+r)r.</math>
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| Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.<ref>Kodokostas, Dimitrios, "Triangle Equalizers," ''Mathematics Magazine'' 83, April 2010, pp. 141-146.</ref>
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| The distance from the incenter to the [[centroid]] is less than one third the length of the longest median of the triangle.<ref name=Franzsen>[http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf Franzsen, William N.. "The distance from the incenter to the Euler line", ''Forum Geometricorum'' 11 (2011): 231–236.]</ref>
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| Denoting the distance from the incenter to the Euler line as ''d'', the length of the longest median as ''v'', the length of the longest side as ''u'', and the semiperimeter as ''s'', the following inequalities hold:<ref name=Franzsen/>
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| :<math>\frac{d}{s} < \frac{d}{u} < \frac{d}{v} < \frac{1}{3}.</math>
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| Denoting the center of the incircle of triangle ''ABC'' as ''I'', we have<ref>Allaire, Patricia R.; Zhou, Junmin; and Yao, Haishen, "Proving a nineteenth century ellipse identity", ''[[Mathematical Gazette]]'' 96, March 2012, 161-165.</ref>
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| :<math>\frac{\overline{IA} \cdot \overline{IA}}{\overline{CA} \cdot \overline{AB}} + \frac{\overline{IB} \cdot \overline{IB}}{\overline{AB} \cdot \overline{BC}} + \frac{\overline{IC} \cdot \overline{IC}}{\overline{BC} \cdot \overline{CA}} = 1.</math>
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| ==Other excircle properties==
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| The circular [[convex hull|hull]] of the excircles is internally tangent to each of the excircles, and thus is an [[Problem of Apollonius|Apollonius circle]].<ref>[http://forumgeom.fau.edu/FG2002volume2/FG200222.pdf Grinberg, Darij, and Yiu, Paul, "The Apollonius Circle as a Tucker Circle", ''Forum Geometricorum'' 2, 2002: pp. 175-182.]</ref> The radius of this Apollonius circle is <math>\frac{r^2+s^2}{4r}</math> where ''r'' is the incircle radius and ''s'' is the semiperimeter of the triangle.<ref>[http://forumgeom.fau.edu/FG2003volume3/FG200320.pdf Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", ''Forum Geometricorum'' 3, 2003, 187-195.]</ref>
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| The following relations hold among the inradius ''r'', the circumradius ''R'', the semiperimeter ''s'', and the excircle radii ''r''<sub>'a''</sub>, ''r''<sub>''b''</sub>, ''r''<sub>''c''</sub>:<ref name=Bell>[http://forumgeom.fau.edu/FG2006volume6/FG200639.pdf Bell, Amy, "Hansen’s right triangle theorem, its converse and a generalization", ''Forum Geometricorum'' 6, 2006, 335–342.]</ref>
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| :<math>r_a+r_b+r_c=4R+r,</math>
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| :<math>r_a r_b+r_br_c+r_cr_a = s^2,</math>
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| :<math>r_a^2 + r_b^2 + r_c^2 = (4R+r)^2 -2s^2,</math>
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| The circle through the centers of the three excircles has radius 2''R''.<ref name=Bell/>
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| If ''H'' is the [[orthocenter]] of triangle ''ABC'', then<ref name=Bell/> | |
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| :<math>r_a+r_b+r_c+r=AH+BH+CH+2R,</math>
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| :<math>r_a^2+r_b^2+r_c^2+r^2=AH^2+BH^2+CH^2+(2R)^2.</math>
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| ==Incircle in a quadrilateral==
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| Some (but not all) [[quadrilateral]]s have an incircle. These are called [[tangential quadrilateral]]s. Among their many properties perhaps the most important is that their opposite sides have equal sums. This is called the [[Pitot theorem]].
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| == See also ==
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| *[[Altitude (triangle)]]
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| *[[Circumscribed circle]]
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| *[[Ex-tangential quadrilateral]]
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| *[[Harcourt's theorem]]
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| *[[Inscribed sphere]]
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| *[[Power of a point]]
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| *[[Steiner inellipse]]
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| *[[Tangential quadrilateral]]
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| *[[Triangle center]]
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| ==References==
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| {{reflist}}
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| *Clark Kimberling, "Triangle Centers and Central Triangles," ''Congressus Numerantium'' 129 (1998) i-xxv and 1-295.
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| *Sándor Kiss, "The Orthic-of-Intouch and Intouch-of-Orthic Triangles," ''Forum Geometricorum'' 6 (2006) 171-177.
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| ==External links==
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| * [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Derivation of formula for radius of incircle of a triangle]
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| * {{MathWorld |title=Incircle |urlname=Incircle}}
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| ===Interactive===
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| *[http://www.mathopenref.com/triangleincenter.html Triangle incenter] [http://www.mathopenref.com/triangleincircle.html Triangle incircle] [http://www.mathopenref.com/polygonincircle.html Incircle of a regular polygon] With interactive animations
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| *[http://www.mathopenref.com/constincircle.html Constructing a triangle's incenter / incircle with compass and straightedge] An interactive animated demonstration
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| * [http://www.cut-the-knot.org/Curriculum/Geometry/AdjacentIncircles.shtml Equal Incircles Theorem] at [[cut-the-knot]]
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| * [http://www.cut-the-knot.org/Curriculum/Geometry/FourIncircles.shtml Five Incircles Theorem] at [[cut-the-knot]]
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| * [http://www.cut-the-knot.org/Curriculum/Geometry/IncirclesInQuadri.shtml Pairs of Incircles in a Quadrilateral] at [[cut-the-knot]]
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| *[http://www.uff.br/trianglecenters/X0001.html An interactive Java applet for the incenter]
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| {{DEFAULTSORT:Incircle And Excircles Of A Triangle}}
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| [[Category:Circles]]
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| [[Category:Triangle geometry]]
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| [[de:Inkreis]]
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