# Altitude (triangle)

"Orthocenter" and "Orthocentre" redirect here. For the orthocentric system, see Orthocentric system.
File:Triangle.Orthocenter.svg
Three altitudes intersecting at the orthocenter

In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e. forming a right angle with) a line containing the base (the opposite side of the triangle). This line containing the opposite side is called the extended base of the altitude. The intersection between the extended base and the altitude is called the foot of the altitude. The length of the altitude, often simply called the altitude, is the distance between the extended base and the vertex. The process of drawing the altitude from the vertex to the foot is known as dropping the altitude of that vertex. It is a special case of orthogonal projection.

Altitudes can be used to compute the area of a triangle: one half of the product of an altitude's length and its base's length equals the triangle's area. Thus the longest altitude is perpendicular to the shortest side of the triangle. The altitudes are also related to the sides of the triangle through the trigonometric functions.

File:Rtriangle.svg
A right triangle, in which the altitude from each acute angle coincides with a leg and intersects the opposite side at (has its foot at) the right-angled vertex.

In an isosceles triangle (a triangle with two congruent sides), the altitude having the incongruent side as its base will have the midpoint of that side as its foot. Also the altitude having the incongruent side as its base will form the angle bisector of the vertex.

It is common to mark the altitude with the letter h (as in height), often subscripted with the name of the side the altitude comes from.

In a right triangle, the altitude with the hypotenuse c as base divides the hypotenuse into two lengths p and q. If we denote the length of the altitude by hc, we then have the relation

$h_c=\sqrt{pq}$  (Geometric mean theorem)

For acute and right triangles the feet of the altitudes all fall on the triangle's interior or edge. In an obtuse triangle (one with an obtuse angle), the foot of the altitude to the obtuse-angled vertex falls on the opposite side, but the feet of the altitudes to the acute-angled vertices fall on the opposite extended side, exterior to the triangle. This is illustrated in the diagram to the right: in this obtuse triangle, an altitude dropped perpendicularly from the top vertex, which has an acute angle, intersects the extended horizontal side outside the triangle.

## Orthocenter

File:Triangle.Orthocenter.svg
Three altitudes intersecting at the orthocenter

The three altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle if and only if the triangle is acute (i.e. does not have an angle greater than or equal to a right angle). If one angle is a right angle, the orthocenter coincides with the vertex of the right angle.

The product of the distances from the orthocenter to a vertex and to the foot of the corresponding altitude is the same for all three altitudes.:p. 176 This product is the squared radius of the triangle's polar circle.

The orthocenter H, the centroid G, the circumcenter O, and the center N 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 between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter:

$OH=2NH,$
$2OG=GH.$

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

$HI < HG,$
$HG > IG.$

In terms of the sides a, b, c, inradius r and circumradius R,

$OH^2 = R^2 -8R^2 \cos A \cos B \cos C=9R^2-(a^2+b^2+c^2),$ :p. 449
$HI^2 = 2r^2 -4R^2 \cos A \cos B \cos C.$

The isogonal conjugate and also the complement of the orthocenter is the circumcenter.

Four points in the plane such that one of them is the orthocenter of the triangle formed by the other three are called an orthocentric system or orthocentric quadrangle.

Let A, B, C denote the angles of the reference triangle, and let a = |BC|, b = |CA|, c = |AB| be the sidelengths. The orthocenter has trilinear coordinates

$\sec A:\sec B:\sec C = \cos A-\sin B \sin C:\cos B-\sin C \sin A:\cos C-\sin A\sin B,$

$\displaystyle (a^2+b^2-c^2)(a^2-b^2+c^2) : (a^2+b^2-c^2)(-a^2+b^2+c^2) : (a^2-b^2+c^2)(-a^2+b^2+c^2)$
$=\tan A:\tan B:\tan C.$

Denote the vertices of a triangle as A, B, and C and the orthocenter as H, and let D, E, and F denote the feet of the altitudes from A, B, and C respectively. Then:

• 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: (This property and the next one are applications of a more general property of any interior point and the three cevians through it.)
$\frac{HD}{AD} + \frac{HE}{BE} + \frac{HF}{CF} = 1.$
• 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:
$\frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 2.$
• The product of the lengths of the segments that the orthocenter divides an altitude into is the same for all three altitudes:
$AH \cdot HD = BH \cdot HE = CH \cdot HF.$
• If any altitude, say AD, is extended to intersect the circumcircle at P, so that AP is a chord of the circumcircle, then the foot D bisects segment HP:
$HD = DP.$

Denote the orthocenter of triangle ABC as H, denote the sidelengths as a, b, and c, and denote the circumradius of the triangle as R. Then

$a^2+b^2+c^2+AH^2+BH^2+CH^2 = 12R^2.$

In addition, denoting r as the radius of the triangle's incircle, ra, rb, and rc as the radii if its excircles, and R again as the radius of its circumcircle, the following relations hold regarding the distances of the orthocenter from the vertices:

$r_a+r_b+r_c+r=AH+BH+CH+2R,$
$r_a^2+r_b^2+r_c^2+r^2=AH^2+BH^2+CH^2+(2R)^2.$

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.

A circumconic passing through the orthocenter of a triangle is a rectangular hyperbola.

In any acute triangle, the inscribed triangle with the smallest perimeter is the pedal triangle of the orthocenter (the triangle whose vertices are the feet of the perpendiculars from the orthocenter to the sides).:p.168 The sides of the pedal triangle of the orthocenter are parallel to the tangents to the circumcircle at the original triangle's vertices.:p.172

## Orthic triangle

File:Altitudes and orthic triangle SVG.svg
Triangle abc is the orthic triangle of triangle ABC

If the triangle ABC is oblique (not right-angled), the points of intersection of the altitudes with the sides of the triangle form another triangle, A'B'C', called the orthic triangle or altitude triangle. It is the pedal triangle of the orthocenter of the original triangle. Also, the incenter (that is, the center for the inscribed circle) of the orthic triangle is the orthocenter of the original triangle.

The sides of the orthic triangle meet the sides of its reference triangle at three collinear points.:p. 165

The orthic triangle is closely related to the tangential triangle, constructed as follows: let LA be the line tangent to the circumcircle of triangle ABC at vertex A, and define LB and LC analogously. Let A" = LB ∩ LC, B" = LC ∩ LA, C" = LC ∩ LA. The tangential triangle is A"B"C", whose sides are the tangents to the reference triangle's circumcircle at its vertices; it is homothetic to the orthic triangle. The circumcenter of the tangential triangle, and the center of similitude of the orthic and tangential triangles, are on the Euler line.:p. 447

The orthic triangle provides the solution to Fagnano's problem, posed in 1775, of finding for the minimum perimeter triangle inscribed in a given acute-angle triangle.

The orthic triangle of an acute triangle gives a triangular light route.

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

• A' = 0 : sec B : sec C
• B' = sec A : 0 : sec C
• C' = sec A : sec B : 0

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

• A" = −a : b : c
• B" = a : −b : c
• C" = a : b : −c

### Altitude in terms of the sides

For any triangle with sides a, b, c and semiperimeter s = (a+b+c) / 2, the altitude from side a is given by

$h_a=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}.$

This follows from combining Heron's formula for the area of a triangle in terms of the sides with the area formula (1/2)×base×height, where the base is taken as side a and the height is the altitude from a.

Consider an arbitrary triangle with sides a, b, c and with corresponding altitudes ha, hb, and hc. The altitudes and the incircle radius r are related by

$\displaystyle \frac{1}{r}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}.$

Denoting the altitude from one side of a triangle as ha, the other two sides as b and c, and the triangle's circumradius (radius of the triangle's circumscribed circle) as R, the altitude is given by:p. 71

$h_a=\frac{bc}{2R}.$

### Interior point

If p1, p2, and p3 are the perpendicular distances from any point P to the sides, and h1, h2, and h3 are the altitudes to the respective sides, then:p. 74

$\frac{p_1}{h_1} +\frac{p_2}{h_2} + \frac{p_3}{h_3} = 1.$

### Area theorem

Denoting the altitudes of any triangle from sides a, b, and c respectively as $h_a$, $h_b$, and $h_c$,and denoting the semi-sum of the reciprocals of the altitudes as $H = (h_a^{-1} + h_b^{-1} + h_c^{-1})/2$ we have

$\mathrm{Area}^{-1} = 4 \sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.$

### General point on an altitude

If E is any point on an altitude AD of any triangle ABC, then:77–78

$AC^2+EB^2=AB^2+CE^2.$

### Feet of the altitudes

The lines connecting the feet of the altitudes intersect the opposite sides at collinear points.:p.199

### Special case triangles

#### Equilateral triangle

For any point P within an equilateral triangle, the sum of the perpendiculars to the three sides is equal to the altitude of the triangle. This is Viviani's theorem.

#### Right triangle

In a right triangle the three altitudes ha, hb, and hc (the first two of which equal the leg lengths b and a respectively) are related according to

$\frac{1}{h_a ^2}+\frac{1}{h_b ^2}=\frac{1}{h_c ^2}.$