The orthocenter of a triangle, usually denoted by, is the point where the three (possibly extended) altitudes intersect. The orthocenter lies inside the triangle if and only if the triangle is acute. For a right triangle, the orthocenter coincides with the vertex at the right angle.
Let denote the vertices and also the angles of the triangle, and let
a=\left|\overline{BC}\right|,b=\left|\overline{CA}\right|,c=\left|\overline{AB}\right|
and barycentric 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 an acute triangle's interior, on the right-angled vertex of a right triangle, and exterior to an obtuse triangle.
In the complex plane, let the points represent the numbers and assume that the circumcenter of triangle is located at the origin of the plane. Then, the complex number
zH=zA+zB+zC
is represented by the point, namely the altitude of triangle .[2] From this, the following characterizations of the orthocenter by means of free vectors can be established straightforwardly:
\vec{OH}=\sum\limits\scriptstyle\rm\vec{OA}, 2 ⋅ \vec{HO}=\sum\limits\scriptstyle\rm\vec{HA}.
The first of the previous vector identities is also known as the problem of Sylvester, proposed by James Joseph Sylvester.[3]
Let denote the feet of the altitudes from respectively. Then:
\overline{AH} ⋅ \overline{HD}=\overline{BH} ⋅ \overline{HE}=\overline{CH} ⋅ \overline{HF}.
The circle centered at having radius the square root of this constant is the triangle's polar circle.
| \overline{HD | |
| \overline{AH | |
Denote the circumradius of the triangle by . Then[7]
a2+b2+c2+\overline{AH}2+\overline{BH}2+\overline{CH}2=12R2.
In addition, denoting as the radius of the triangle's incircle, as the radii of its excircles, and again as the radius of its circumcircle, the following relations hold regarding the distances of the orthocenter from the vertices:[8]
\begin{align} &ra+rb+rc+r=\overline{AH}+\overline{BH}+\overline{CH}+2R,\\ &
| 2 | |
| r | |
| a |
+
| 2 | |
| r | |
| b |
+
| 2 | |
| r | |
| c |
+r2=\overline{AH}2+\overline{BH}2+\overline{CH}2+(2R)2. \end{align}
If any altitude, for example,, is extended to intersect the circumcircle at, so that is a chord of the circumcircle, then the foot bisects segment :[4]
\overline{HD}=\overline{DP}.
The directrices of all parabolas that are externally tangent to one side of a triangle and tangent to the extensions of the other sides pass through the orthocenter.[9]
A circumconic passing through the orthocenter of a triangle is a rectangular hyperbola.[10]
See main article: Nine-point circle. The orthocenter, the centroid, the circumcenter, and the center of the nine-point circle all lie on a single line, known as the Euler line. The center of the nine-point circle lies at the midpoint 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:
\begin{align} &\overline{OH}=2\overline{NH},\\ &2\overline{OG}=\overline{GH}. \end{align}
The orthocenter is closer to the incenter than it is to the centroid, and the orthocenter is farther than the incenter is from the centroid:
\begin{align} \overline{HI}&<\overline{HG},\\ \overline{HG}&>\overline{IG}. \end{align}
In terms of the sides,,, inradius and circumradius,[11] [12]
\begin{align} \overline{OH}2&=R2-8R2\cosA\cosB\cosC\\ &=9R2-(a2+b2+c2),\\ \overline{HI}2&=2r2-4R2\cosA\cosB\cosC. \end{align}
If the triangle is oblique (does not contain a right-angle), the pedal triangle of the orthocenter of the original triangle is called the orthic triangle or altitude triangle. That is, the feet of the altitudes of an oblique triangle form the orthic triangle, . Also, the incenter (the center of the inscribed circle) of the orthic triangle is the orthocenter of the original triangle .[13]
Trilinear coordinates for the vertices of the orthic triangle are given by
The extended sides of the orthic triangle meet the opposite extended sides of its reference triangle at three collinear points.
In any acute triangle, the inscribed triangle with the smallest perimeter is the orthic triangle. This is the solution to Fagnano's problem, posed in 1775. The sides of the orthic triangle are parallel to the tangents to the circumcircle at the original triangle's vertices.
The orthic triangle of an acute triangle gives a triangular light route.[14]
The tangent lines of the nine-point circle at the midpoints of the sides of are parallel to the sides of the orthic triangle, forming a triangle similar to the orthic triangle.
The orthic triangle is closely related to the tangential triangle, constructed as follows: let be the line tangent to the circumcircle of triangle at vertex, and define analogously. Let
A''=LB\capLC,
B''=LC\capLA,
C''=LC\capLA.
Trilinear coordinates for the vertices of the tangential triangle are given byThe reference triangle and its orthic triangle are orthologic triangles.
For more information on the orthic triangle, see here.
The theorem that the three altitudes of a triangle concur (at the orthocenter) is not directly stated in surviving Greek mathematical texts, but is used in the Book of Lemmas (proposition 5), attributed to Archimedes (3rd century BC), citing the "commentary to the treatise about right-angled triangles", a work which does not survive. It was also mentioned by Pappus (Mathematical Collection, VII, 62; 340).[15] The theorem was stated and proved explicitly by al-Nasawi in his (11th century) commentary on the Book of Lemmas, and attributed to al-Quhi .[16]
This proof in Arabic was translated as part of the (early 17th century) Latin editions of the Book 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 his Geometrie (1619), and Isaac Newton proved it in an unfinished treatise Geometry of Curved Lines Later William Chapple proved it in 1749.[17]
A particularly elegant proof is due to François-Joseph Servois (1804) and independently Carl 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 the medial triangle of the new triangle, and the altitudes of the original triangle are the perpendicular bisectors of the new triangle, and therefore concur (at the circumcenter of the new triangle).[18]