The parameters in a triangle inequality can be the side lengths, the semiperimeter, the angle measures, the values of trigonometric functions of those angles, the area of the triangle, the medians of the sides, the altitudes, the lengths of the internal angle bisectors from each angle to the opposite side, the perpendicular bisectors of the sides, the distance from an arbitrary point to another point, the inradius, the exradii, the circumradius, and/or other quantities.
While all of the above inequalities are true because a, b, and c must follow the basic triangle inequality that the longest side is less than half the perimeter, the following relations hold for all positive a, b, and c:[1]: p.267 each holding with equality only when a = b = c. This says that in the non-equilateral case the harmonic mean of the sides is less than their geometric mean, which in turn is less than their arithmetic mean, and which in turn is less than their quadratic mean.
Weitzenböck's inequality is, in terms of area T,[1]: p. 290 with equality only in the equilateral case.
[10] This is strengthened by Bonnesen's inequality also strengthens the isoperimetric inequality: We also have with equality only in the equilateral case; for semiperimeter s; and Ono's inequality for acute triangles (those with all angles less than 90°) is The area of the triangle can be compared to the area of the incircle: with equality only for the equilateral triangle.
of a triangle each connect a vertex with the midpoint of the opposite side, and the sum of their lengths satisfies[1]: p. 271 Moreover,[2]: p.12, #589 with equality only in the equilateral case, and for inradius r,[2]: p.22, #846 If we further denote the lengths of the medians extended to their intersections with the circumcircle as Ma , Mb , and Mc , then[2]: p.16, #689 The centroid G is the intersection of the medians.
Let AG, BG, and CG meet the circumcircle at U, V, and W respectively.
Denoting as IA, IB, IC the distances of the incenter from the vertices, the following holds:[2]: p.192, #339.3 The three medians of any triangle can form the sides of another triangle:[13]: p. 592 Furthermore,[14]: Coro.
then[2]: 222, #67 We also have[2]: p.140, #3150 For internal angle bisectors ta, tb, tc from vertices A, B, C and circumcenter R and incenter r, we have[2]: p.125, #3005 The reciprocals of the altitudes of any triangle can themselves form a triangle:[15] The internal angle bisectors are segments in the interior of the triangle reaching from one vertex to the opposite side and bisecting the vertex angle into two equal angles.
271–3 Further,[2]: p.224, #132 in terms of the medians, and[2]: p.125, #3005 in terms of the altitudes, inradius r and circumradius R. Let Ta , Tb , and Tc be the lengths of the angle bisectors extended to the circumcircle.
[2]: p.20, #795 For incenter I (the intersection of the internal angle bisectors),[2]: p.127, #3033 For midpoints L, M, N of the sides,[2]: p.152, #J53 For incenter I, centroid G, circumcenter O, nine-point center N, and orthocenter H, we have for non-equilateral triangles the distance inequalities[16]: p.232 and and we have the angle inequality[16]: p.233 In addition,[16]: p.233, Lemma 3 where v is the longest median.
Three triangles with vertex at the incenter, OIH, GIH, and OGI, are obtuse:[16]: p.232 Since these triangles have the indicated obtuse angles, we have and in fact the second of these is equivalent to a result stronger than the first, shown by Euler:[17][18] The larger of two angles of a triangle has the shorter internal angle bisector:[19]: p.72, #114 These inequalities deal with the lengths pa etc.
of the triangle-interior portions of the perpendicular bisectors of sides of the triangle.
is the shortest side of the triangle, then We also have Ptolemy's inequality[2]: p.19, #770 for interior point P and likewise for cyclic permutations of the vertices.
More strongly, Barrow's inequality states that if the interior bisectors of the angles at interior point P (namely, of ∠APB, ∠BPC, and ∠CPA) intersect the triangle's sides at U, V, and W, then[23] Also stronger than the Erdős–Mordell inequality is the following:[24] Let D, E, F be the orthogonal projections of P onto BC, CA, AB respectively, and H, K, L be the orthogonal projections of P onto the tangents to the triangle's circumcircle at A, B, C respectively.
Then With orthogonal projections H, K, L from P onto the tangents to the triangle's circumcircle at A, B, C respectively, we have[25] where R is the circumradius.
Again with distances PD, PE, PF of the interior point P from the sides we have these three inequalities:[2]: p.29, #1045 For interior point P with distances PA, PB, PC from the vertices and with triangle area T,[2]: p.37, #1159 and[2]: p.26, #965 For an interior point P, centroid G, midpoints L, M, N of the sides, and semiperimeter s,[2]: p.140, #3164 [2]: p.130, #3052 Moreover, for positive numbers k1, k2, k3, and t with t less than or equal to 1:[26]: Thm.1 while for t > 1 we have[26]: Thm.2 There are various inequalities for an arbitrary interior or exterior point in the plane in terms of the radius r of the triangle's inscribed circle.
Furthermore, for circumradius R, Let ABC be a triangle, let G be its centroid, and let D, E, and F be the midpoints of BC, CA, and AB, respectively.
For any point P in the plane of ABC: The Euler inequality for the circumradius R and the inradius r states that with equality only in the equilateral case.[31]: p.
198 A stronger version[5]: p. 198 is By comparison,[2]: p.183, #276.2 where the right side could be positive or negative.
In the latter double inequality, the first part holds with equality if and only if the triangle is isosceles with an apex angle of at least 60°, and the last part holds with equality if and only if the triangle is isosceles with an apex angle of at most 60°.
The circumcenter is inside the incircle if and only if[32] Further, Blundon's inequality states that[5]: p. 206, [33][34] We also have, for all acute triangles,[35] For incircle center I, let AI, BI, and CI extend beyond I to intersect the circumcircle at D, E, and F respectively.
Moreover, for circumcenter O, let lines AO, BO, and CO intersect the opposite sides BC, CA, and AB at U, V, and W respectively.
The circumradius is at least twice the distance between the first and second Brocard points B1 and B2:[38] For the inradius r we have[1]: pp.
We additionally have and The exradii and medians are related by[2]: p.66, #1680 In addition, for an acute triangle the distance between the incircle center I and orthocenter H satisfies[2]: p.26, #954 with the reverse inequality for an obtuse triangle.
If one of these squares has side length xa and another has side length xb with xa < xb, then[39]: p. 115 Moreover, for any square inscribed in any triangle we have[2]: p.18, #729 [39] A triangle's Euler line goes through its orthocenter, its circumcenter, and its centroid, but does not go through its incenter unless the triangle is isosceles.
[16]: p.231 For all non-isosceles triangles, the distance d from the incenter to the Euler line satisfies the following inequalities in terms of the triangle's longest median v, its longest side u, and its semiperimeter s:[16]: p. 234, Propos.5 For all of these ratios, the upper bound of 1/3 is the tightest possible.
[16]: p.235, Thm.6 In right triangles the legs a and b and the hypotenuse c obey the following, with equality only in the isosceles case:[1]: p. 280 In terms of the inradius, the hypotenuse obeys[1]: p. 281 and in terms of the altitude from the hypotenuse the legs obey[1]: p. 282 If the two equal sides of an isosceles triangle have length a and the other side has length c, then the internal angle bisector t from one of the two equal-angled vertices satisfies[2]: p.169, #
A triangle is equilateral if and only if, for every point P in the plane, with distances PD, PE, and PF to the triangle's sides and distances PA, PB, and PC to its vertices,[2]: p.178, #235.4
That is, in triangles ABC and DEF with sides a, b, c, and d, e, f respectively (with a opposite A etc.
), if a = d and b = e and angle C > angle F, then The converse also holds: if c > f, then C > F. The angles in any two triangles ABC and DEF are related in terms of the cotangent function according to[6] In a triangle on the surface of a sphere, as well as in elliptic geometry, This inequality is reversed for hyperbolic triangles.