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Altitude (triangle)

From Wikipedia, the free encyclopedia
Perpendicular line segment from a triangle's side to opposite vertex
The altitude from A (dashed line segment) intersects the extended base at D (a point outside the triangle).

Ingeometry, analtitude of atriangle is aline segment through a givenvertex (calledapex) andperpendicular to aline containing the side oredge opposite the apex. This (finite) edge and (infinite) line extension are called, respectively, thebase andextended base of the altitude. Thepoint at the intersection of the extended base and the altitude is called thefoot of the altitude. The length of the altitude, often simply called "the altitude" or "height", symbolh, is the distance between the foot and the apex. The process of drawing the altitude from a vertex to the foot is known asdropping the altitude at that vertex. It is a special case oforthogonal projection.

Altitudes can be used in the computation of thearea of a triangle: one-half of the product of an altitude's length and its base's length (symbolb) equals the triangle's area:A=hb/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 thetrigonometric functions.

In anisosceles triangle (a triangle with twocongruent sides), the altitude having the incongruent side as its base will have themidpoint of that side as its foot. Also the altitude having the incongruent side as its base will be theangle bisector of the vertex angle.

In aright triangle, the altitude drawn to thehypotenusec divides the hypotenuse into two segments of lengthsp andq. If we denote the length of the altitude byhc, we then have the relation

hc=pq{\displaystyle h_{c}={\sqrt {pq}}}  (geometric mean theorem; seespecial cases,inverse Pythagorean theorem)
In a right triangle, the altitude from each acute angle coincides with a leg and intersects the opposite side at (has its foot at) the right-angled vertex, which is the orthocenter.

For acute triangles, the feet of the altitudes all fall on the triangle's sides (not extended). In an obtuse triangle (one with anobtuse 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 oppositeextended 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

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The geometric altitude figures prominently in many important theorems and their proofs. For example, besides those theorems listed below, the altitude plays a central role in proofs of both theLaw of sines andLaw of cosines.

Orthocenter

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This section is an excerpt fromOrthocenter.[edit]
The three altitudes of a triangle intersect at the orthocenter, which for anacute triangle is inside the triangle.
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] For anequilateral triangle, alltriangle centers (including the orthocenter) coincide at itscentroid.

Altitude in terms of the sides

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For any triangle with sidesa, b, c andsemiperimeters=12(a+b+c),{\displaystyle s={\tfrac {1}{2}}(a+b+c),} the altitude from sidea (the base) is given by

ha=2s(sa)(sb)(sc)a.{\displaystyle h_{a}={\frac {2{\sqrt {s(s-a)(s-b)(s-c)}}}{a}}.}

This follows from combiningHeron's formula for the area of a triangle in terms of the sides with the area formula12×base×height,{\displaystyle {\tfrac {1}{2}}\times {\text{base}}\times {\text{height}},} where the base is taken as sidea and the height is the altitude from the vertexA (opposite sidea).

By exchanginga withb orc, this equation can also used to find the altitudeshb andhc, respectively.

Any two altitudes of a triangle are inversely proportional with the sides on which they fall.

Inradius theorems

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Consider an arbitrary triangle with sidesa, b, c and with correspondingaltitudesha, hb, hc. The altitudes and theincircle radiusr are related by[3]: Lemma 1 

1r=1ha+1hb+1hc.{\displaystyle \displaystyle {\frac {1}{r}}={\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}}.}

Circumradius theorem

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Denoting the altitude from one side of a triangle asha, the other two sides asb andc, and the triangle'scircumradius (radius of the triangle's circumscribed circle) asR, the altitude is given by[4]

ha=bc2R.{\displaystyle h_{a}={\frac {bc}{2R}}.}

Interior point

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Ifp1,p2,p3 are the perpendicular distances from any pointP to the sides, andh1,h2,h3 are the altitudes to the respective sides, then[5]

p1h1+p2h2+p3h3=1.{\displaystyle {\frac {p_{1}}{h_{1}}}+{\frac {p_{2}}{h_{2}}}+{\frac {p_{3}}{h_{3}}}=1.}

Area theorem

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Denoting the altitudes of any triangle from sidesa, b, c respectively asha, hb, hc, and the semi-sum of the reciprocals of the altitudes asH=12(ha1+hb1+hc1){\displaystyle \textstyle H={\tfrac {1}{2}}(h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1})} then the reciprocal of area is[6]

Area1=4H(Hha1)(Hhb1)(Hhc1).{\displaystyle \mathrm {Area} ^{-1}=4{\sqrt {H(H-h_{a}^{-1})(H-h_{b}^{-1})(H-h_{c}^{-1})}}.}

General point on an altitude

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IfE is any point on an altitudeAD of any triangleABC, then[7]: 77–78 

AC¯2+EB¯2=AB¯2+CE¯2.{\displaystyle {\overline {AC}}^{2}+{\overline {EB}}^{2}={\overline {AB}}^{2}+{\overline {CE}}^{2}.}

Triangle inequality

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Since the area of the triangle is12aha=12bhb=12chc{\displaystyle {\tfrac {1}{2}}ah_{a}={\tfrac {1}{2}}bh_{b}={\tfrac {1}{2}}ch_{c}}, the triangle inequalitya<b+c{\displaystyle a<b+c} implies[8]

1ha<1hb+1hc{\displaystyle {\frac {1}{h_{a}}}<{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}}}.

Special cases

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Equilateral triangle

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From any pointP within anequilateral triangle, the sum of the perpendiculars to the three sides is equal to the altitude of the triangle. This isViviani's theorem.

Right triangle

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The altitude of a right triangle from its right angle to its hypotenuse is the geometric mean of the lengths of the segments the hypotenuse is split into. UsingPythagoras' theorem on the 3 triangles of sides(p + q,r,s ),(r,p,h ) and(s,h,q ),
(p+q)2=r2+s2p2+2pq+q2=p2+h2+h2+q22pq=2h2h=pq{\displaystyle {\begin{aligned}(p+q)^{2}\;\;&=\quad r^{2}\;\;\,+\quad s^{2}\\p^{2}\!\!+\!2pq\!+\!q^{2}&=\overbrace {p^{2}\!\!+\!h^{2}} +\overbrace {h^{2}\!\!+\!q^{2}} \\2pq\quad \;\;\;&=2h^{2}\;\therefore h\!=\!{\sqrt {pq}}\\\end{aligned}}}
Comparison of the inverse Pythagorean theorem with the Pythagorean theorem

In a right triangle with legsa andb and hypotenusec, each of the legs is also an altitude:ha=b{\displaystyle h_{a}=b} andhb=a{\displaystyle h_{b}=a}. The third altitude can be found by the relation[9][10]

1hc2=1ha2+1hb2=1a2+1b2.{\displaystyle {\frac {1}{h_{c}^{2}}}={\frac {1}{h_{a}^{2}}}+{\frac {1}{h_{b}^{2}}}={\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}.}

This is also known as theinverse Pythagorean theorem.

Note in particular:

12ACBC=12ABCDCD=ACBCAB{\displaystyle {\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\[4pt]CD&={\frac {AC\cdot BC}{AB}}\\[4pt]\end{aligned}}}

See also

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Notes

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  1. ^Smart 1998, p. 156
  2. ^abBerele & Goldman 2001, p. 118
  3. ^Andrica, Dorin; Marinescu, Dan Ştefan (2017)."New Interpolation Inequalities to Euler's R ≥ 2r"(PDF).Forum Geometricorum.17:149–156. Archived fromthe original(PDF) on 2018-04-24.
  4. ^Johnson 2007, p. 71, Section 101a
  5. ^Johnson 2007, p. 74, Section 103c
  6. ^Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle,"Mathematical Gazette 89, November 2005, 494.
  7. ^Alfred S. Posamentier and Charles T. Salkind,Challenging Problems in Geometry, Dover Publishing Co., second revised edition, 1996.
  8. ^Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle",Mathematical Gazette 89 (November 2005), 494.
  9. ^Voles, Roger, "Integer solutions ofa2+b2=d2{\displaystyle a^{-2}+b^{-2}=d^{-2}},"Mathematical Gazette 83, July 1999, 269–271.
  10. ^Richinick, Jennifer, "The upside-down Pythagorean Theorem,"Mathematical Gazette 92, July 2008, 313–317.

References

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External links

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