Theorthocenter of atriangle, usually denoted byH, is thepoint where the three (possibly extended)altitudes intersect.^[1][2] The orthocenter lies inside the triangleif and only if the triangle isacute. For aright triangle, the orthocenter coincides with thevertex at the right angle.^[2]

LetA, B, C denote the vertices and also the angles of the triangle, and let${\displaystyle a=\left|\overline{BC}\right|,b=\left|\overline{CA}\right|,c=\left|\overline{AB}\right|}$ be the side lengths. The orthocenter hastrilinear coordinates^[3]

andbarycentric coordinates

Since barycentric coordinates are all positive for a point in a triangle's interior but at least one is negative for a point in the exterior, and two of the barycentric coordinates are zero for a vertex point, the barycentric coordinates given for the orthocenter show that the orthocenter is in anacute triangle's interior, on the right-angled vertex of aright triangle, and exterior to anobtuse triangle.

In thecomplex plane, let the pointsA, B, C represent thenumbersz_A, z_B, z_C and assume that thecircumcenter of triangle△ABC is located at the origin of the plane. Then, the complex number

${\displaystyle z_H=z_A+z_B+z_C}$

is represented by the pointH, namely the altitude of triangle△ABC.^[4] From this, the following characterizations of the orthocenterH by means offree vectors can be established straightforwardly:

${\displaystyle \overrightarrow{OH}=\sum _{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{i}\mathrm{c}}\overrightarrow{OA},2\cdot \overrightarrow{HO}=\sum _{\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{i}\mathrm{c}}\overrightarrow{HA}.}$

The first of the previous vector identities is also known as theproblem of Sylvester, proposed byJames Joseph Sylvester.^[5]

LetD, E, F denote the feet of the altitudes fromA, B, C respectively. Then:

• The product of the lengths of the segments that the orthocenter divides an altitude into is the same for all three altitudes:^[6][7]

${\displaystyle \overline{AH}\cdot \overline{HD}=\overline{BH}\cdot \overline{HE}=\overline{CH}\cdot \overline{HF}.}$

The circle centered atH having radius the square root of this constant is the triangle'spolar circle.^[8]

• The sum of the ratios on the three altitudes of the distance of the orthocenter from the base to the length of the altitude is 1:^[9] (This property and the next one are applications of amore general property of any interior point and the threecevians through it.)

${\displaystyle \frac{\overline{HD}}{\overline{AD}}+\frac{\overline{HE}}{\overline{BE}}+\frac{\overline{HF}}{\overline{CF}}=1.}$

• The sum of the ratios on the three altitudes of the distance of the orthocenter from the vertex to the length of the altitude is 2:^[9]

${\displaystyle \frac{\overline{AH}}{\overline{AD}}+\frac{\overline{BH}}{\overline{BE}}+\frac{\overline{CH}}{\overline{CF}}=2.}$

• Theisogonal conjugate of the orthocenter is thecircumcenter of the triangle.^[10]
• Theisotomic conjugate of the orthocenter is thesymmedian point of theanticomplementary triangle.^[11]
• Four points in the plane, such that one of them is the orthocenter of the triangle formed by the other three, is called anorthocentric system or orthocentric quadrangle.

This section is an excerpt fromOrthocentric system.[edit]

Ingeometry, an orthocentric system is aset of fourpoints on aplane, one of which is the orthocenter of thetriangle formed by the other three. Equivalently, the lines passing through disjoint pairs among the points areperpendicular, and the four circles passing through any three of the four points have the same radius.^[12]

If four points form an orthocentric system, theneach of the four points is the orthocenter of the other three. These four possible triangles will all have the samenine-point circle. Consequently these four possible triangles must all havecircumcircles with the samecircumradius.

Denote thecircumradius of the triangle byR. Then^[13][14]

${\displaystyle a^2+b^2+c^2+{\overline{AH}}^2+{\overline{BH}}^2+{\overline{CH}}^2=12R^2.}$

In addition, denotingr as the radius of the triangle'sincircle,r_a, r_b, r_c as the radii of itsexcircles, andR again as the radius of its circumcircle, the following relations hold regarding the distances of the orthocenter from the vertices:^[15]

If any altitude, for example,AD, is extended to intersect the circumcircle atP, so thatAD is a chord of the circumcircle, then the footD bisects segmentHP:^[7]

${\displaystyle \overline{HD}=\overline{DP}.}$

Thedirectrices of allparabolas that are externally tangent to one side of a triangle and tangent to the extensions of the other sides pass through the orthocenter.^[16]

Acircumconic passing through the orthocenter of a triangle is arectangular hyperbola.^[17]

The orthocenterH, thecentroidG, thecircumcenterO, and the centerN of thenine-point circle all lie on a single line, known as theEuler line.^[18] The center of the nine-point circle lies at themidpoint of the Euler line, between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half of that between the centroid and the orthocenter:^[19]

The orthocenter is closer to theincenterI than it is to the centroid, and the orthocenter is farther than the incenter is from the centroid:

In terms of the sidesa,b,c,inradiusr andcircumradiusR,^{[20][21]: p. 449}

If the triangle△ABC isoblique (does not contain a right-angle), thepedal triangle of the orthocenter of the original triangle is called theorthic triangle oraltitude triangle. That is, the feet of the altitudes of an oblique triangle form the orthic triangle,△DEF. Also, the incenter (the center of the inscribed circle) of the orthic triangle△DEF is the orthocenter of the original triangle△ABC.^[22]

Trilinear coordinates for the vertices of the orthic triangle are given by

Theextended sides of the orthic triangle meet the opposite extended sides of its reference triangle at threecollinear points.^[23][24][22]

In anyacute triangle, the inscribed triangle with the smallest perimeter is the orthic triangle.^[25] This is the solution toFagnano's problem, posed in 1775.^[26] The sides of the orthic triangle are parallel to the tangents to the circumcircle at the original triangle's vertices.^[27]

The orthic triangle of an acute triangle gives a triangular light route.^[28]

The tangent lines of the nine-point circle at the midpoints of the sides of△ABC are parallel to the sides of the orthic triangle, forming a triangle similar to the orthic triangle.^[29]

The orthic triangle is closely related to thetangential triangle, constructed as follows: letL_A be the line tangent to the circumcircle of triangle△ABC at vertexA, and defineL_B, L_C analogously. Let${\displaystyle A^{″}=L_B\cap L_C,}$${\displaystyle B^{″}=L_C\cap L_A,}$${\displaystyle C^{″}=L_C\cap L_A.}$ The tangential triangle is△A"B"C", whose sides are the tangents to triangle△ABC's circumcircle at its vertices; it ishomothetic to the orthic triangle. The circumcenter of the tangential triangle, and thecenter of similitude of the orthic and tangential triangles, are on theEuler line.^{[21]: p. 447}

Trilinear coordinates for the vertices of the tangential triangle are given byThe reference triangle and its orthic triangle areorthologic triangles.

For more information on the orthic triangle, seehere.

The theorem that the three altitudes of a triangle concur (at the orthocenter) is not directly stated in survivingGreek mathematical texts, but is used in theBook of Lemmas (proposition 5), attributed toArchimedes (3rd century BC), citing the "commentary to the treatise about right-angled triangles", a work which does not survive. It was also mentioned byPappus (Mathematical Collection, VII, 62;c. 340).[30] The theorem was stated and proved explicitly byal-Nasawi in his (11th century) commentary on theBook of Lemmas, and attributed toal-Quhi (fl. 10th century).[31]

This proof in Arabic was translated as part of the (early 17th century) Latin editions of theBook of Lemmas, but was not widely known in Europe, and the theorem was therefore proven several more times in the 17th–19th century.Samuel Marolois proved it in hisGeometrie (1619), andIsaac Newton proved it in an unfinished treatiseGeometry of Curved Lines(c. 1680).[30] LaterWilliam Chapple proved it in 1749.^[32]

A particularly elegant proof is due toFrançois-Joseph Servois (1804) and independentlyCarl Friedrich Gauss (1810): Draw a line parallel to each side of the triangle through the opposite point, and form a new triangle from the intersections of these three lines. Then the original triangle is themedial triangle of the new triangle, and the altitudes of the original triangle are theperpendicular bisectors of the new triangle, and therefore concur (at the circumcenter of the new triangle).^[33]

• Triangle center

• Altshiller-Court, Nathan (2007) [1952],College Geometry, Dover
• Berele, Allan; Goldman, Jerry (2001),Geometry: Theorems and Constructions, Prentice Hall,ISBN 0-13-087121-4
• Johnson, Roger A. (2007) [1960],Advanced Euclidean Geometry, Dover,ISBN 978-0-486-46237-0
• Smart, James R. (1998),Modern Geometries (5th ed.), Brooks/Cole,ISBN 0-534-35188-3

• Weisstein, Eric W."Altitude".MathWorld.
• Orthocenter of a triangle With interactive animation
• Animated demonstration of orthocenter construction Compass and straightedge.
• Fagnano's Problem by Jay Warendorff,Wolfram Demonstrations Project.

• Triangle centers

• CS1: unfit URL
• CS1 French-language sources (fr)
• CS1 German-language sources (de)
• Articles with short description
• Short description is different from Wikidata
• Articles with excerpts