Altitude (triangle)
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In geometry, an altitude of a triangle is a line segment through a given vertex (called apex) and perpendicular to a line containing the side or edge opposite the apex. This (finite) edge and (infinite) line extension are called, respectively, the base and extended base of the altitude. The point at the intersection of the extended base and the altitude is called the foot of the altitude. The length of the altitude, often simply called "the altitude" or "height", symbol Template:Mvar, is the distance between the foot and the apex. The process of drawing the altitude from a vertex to the foot is known as dropping the altitude at that vertex. It is a special case of orthogonal projection.
Altitudes can be used in the computation of the area of a triangle: one-half of the product of an altitude's length and its base's length (symbol Template:Mvar) equals the triangle's area: Template:Mvar=Template:MvarTemplate:Mvar/2. 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 be the angle bisector of the vertex angle.
In a right triangle, the altitude drawn to the hypotenuse Template:Mvar divides the hypotenuse into two segments of lengths Template:Mvar and Template:Mvar. If we denote the length of the altitude by Template:Mvar, we then have the relation
- <math>h_c=\sqrt{pq} </math> (Geometric mean theorem; see Special Cases, inverse Pythagorean theorem)
For acute triangles, the feet of the altitudes all fall on the triangle's sides (not extended). In an obtuse triangle (one with an obtuse angle), the foot of the altitude to the obtuse-angled vertex falls in the interior of 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 adjacent diagram: 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.
Theorems
[edit]Orthocenter
[edit]Altitude in terms of the sides
[edit]For any triangle with sides Template:Mvar and semiperimeter <math>s = \tfrac12(a+b+c),</math> the altitude from side Template:Mvar (the base) is given by
- <math>h_a=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}.</math>
This follows from combining Heron's formula for the area of a triangle in terms of the sides with the area formula <math>\tfrac{1}{2} \times \text{base} \times \text{height},</math> where the base is taken as side Template:Mvar and the height is the altitude from the vertex Template:Mvar (opposite side Template:Mvar).
By exchanging Template:Mvar with Template:Mvar or Template:Mvar, this equation can also used to find the altitudes Template:Mvar and Template:Mvar, respectively.
Inradius theorems
[edit]Consider an arbitrary triangle with sides Template:Mvar and with corresponding altitudes Template:Mvar. The altitudes and the incircle radius Template:Mvar are related by<ref>Template:Cite journal</ref>Template:Rp
- <math>\displaystyle \frac{1}{r}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}.</math>
Circumradius theorem
[edit]Denoting the altitude from one side of a triangle as Template:Mvar, the other two sides as Template:Mvar and Template:Mvar, and the triangle's circumradius (radius of the triangle's circumscribed circle) as Template:Mvar, the altitude is given by<ref>Template:Harvnb</ref>
- <math>h_a=\frac{bc}{2R}.</math>
Interior point
[edit]If Template:Math are the perpendicular distances from any point Template:Mvar to the sides, and Template:Math are the altitudes to the respective sides, then<ref>Template:Harvnb</ref>
- <math>\frac{p_1}{h_1} +\frac{p_2}{h_2} + \frac{p_3}{h_3} = 1.</math>
Area theorem
[edit]Denoting the altitudes of any triangle from sides Template:Mvar respectively as Template:Mvar, and denoting the semi-sum of the reciprocals of the altitudes as <math>H = \tfrac{h_a^{-1} + h_b^{-1} + h_c^{-1}}{2}</math> we have<ref>Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.</ref>
- <math>\mathrm{Area}^{-1} = 4 \sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}.</math>
General point on an altitude
[edit]If Template:Mvar is any point on an altitude Template:Mvar of any triangle Template:Math, then<ref name=Posamentier>Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Geometry, Dover Publishing Co., second revised edition, 1996.</ref>Template:Rp
- <math>\overline{AC}^2 + \overline{EB}^2 = \overline{AB}^2 + \overline{CE}^2.</math>
Triangle inequality
[edit]Since the area of the triangle is <math>\tfrac12 a h_a = \tfrac12 b h_b = \tfrac12 c h_c</math>, the triangle inequality <math>a < b+ c</math> implies<ref>Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle", Mathematical Gazette 89 (November 2005), 494.</ref>
- <math>\frac1{h_a} < \frac1{h_b}+ \frac1{h_c}</math>.
Special cases
[edit]Equilateral triangle
[edit]From any point Template:Mvar 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
[edit]Template:Right angle altitude.svg
In a right triangle with legs Template:Mvar and Template:Mvar and hypotenuse Template:Mvar, each of the legs is also an altitude: Template:Tmath and Template:Tmath. The third altitude can be found by the relation<ref>Voles, Roger, "Integer solutions of <math>a^{-2}+b^{-2}=d^{-2}</math>," Mathematical Gazette 83, July 1999, 269–271.</ref><ref>Richinick, Jennifer, "The upside-down Pythagorean Theorem," Mathematical Gazette 92, July 2008, 313–317.</ref>
- <math>\frac{1}{h_c ^2} = \frac{1}{h_a ^2}+\frac{1}{h_b ^2} = \frac{1}{a^2}+\frac{1}{b^2}.</math>
This is also known as the inverse Pythagorean theorem.
Note in particular:
- <math>\begin{align}
\tfrac{1}{2} AC \cdot BC &= \tfrac{1}{2} AB \cdot CD \\[4pt]
CD &= \tfrac{AC \cdot BC}{AB} \\[4pt]
\end{align}</math>