Altitude (triangle)

"Orthocenter" and "Orthocentre" redirect here. For the orthocentric system, see Orthocentric system.

Three altitudes intersecting at the orthocenter

An altitude is the perpendicular segment from a vertex to its opposite side. In geometry, an altitude of a triangle is a straight line 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 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.

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).

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

$h=\sqrt{pq}$.

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). See also orthocentric system. If one angle is a right angle, the orthocenter coincides with the vertex of the right angle. Thus for acute and right triangles the feet of the altitudes all fall on the triangle.

The orthocenter, along with the centroid, circumcenter and 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 between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter.

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 and barycentric coordinates

$\displaystyle ((a^2+b^2-c^2)(a^2-b^2+c^2) : (a^2+b^2-c^2)(-a^2+b^2+c^2) : (a^2-b^2+c^2)(-a^2+b^2+c^2)).$

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:

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

• 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, r_a, r_b, and r_c 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:

$r_a+r_b+r_c+r=AH+BH+CH+2R,$

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

Orthic triangle

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 orthic triangle is closely related to the tangential triangle, constructed as follows: let L_A be the line tangent to the circumcircle of triangle ABC at vertex A, and define L_B and L_C analogously. Let A" = L_B ∩ L_C, B" = L_C ∩ L_A, C" = L_C ∩ L_A. The tangential triangle, A"B"C", is homothetic to the orthic triangle.

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

Some additional altitude theorems

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.

Equilateral triangle theorem

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.

Inradius theorems

Consider an arbitrary triangle with sides a, b, c and with corresponding altitudes α, β, η. The altitudes and incircle radius r are related by

$\displaystyle \frac{1}{r}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\eta}.$

Let c, h, s be the sides of three squares associated with the right triangle: the square on the hypotenuse, and the triangle's two inscribed squares respectively. The sides of these squares (c>h>s) and the incircle radius r are related by a similar formula:

$\displaystyle \frac{1}{r}=-{\frac{1}{c}}+\frac{1}{h}+\frac{1}{s}.$

Relation among altitudes of right triangle

In a right triangle the three altitudes α, β, η (the first two of which coincide with the legs b and a respectively) are related according to

$\frac{1}{\alpha ^2}+\frac{1}{\beta ^2}=\frac{1}{\eta ^2}.$

Relation among sides of squares on and in a right triangle

Also in the case of the right triangle, the sides c, h, s of the three above-mentioned squares are related to each other by the symphonic theorem, which states that

$\frac{1}{c^2}+\frac{1}{h^2}=\frac{1}{s^2}.$

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

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

See also

• Triangle center
• Median (geometry)

References

• Weisstein, Eric W., "Altitude", MathWorld.

External links

• Orthocenter of a triangle With interactive animation
• Animated demonstration of orthocenter construction Compass and straightedge.
• An interactive Java applet for the orthocenter
• Fagnano's Problem by Jay Warendorff, Wolfram Demonstrations Project.