Geometric mean theorem

Geometric mean theorem area of grey square = area of grey rectangle: {displaystyle h^{2}=pqLeftrightarrow h={sqrt {pq}}} The right triangle altitude theorem or geometric mean theorem is a result in elementary geometry that describes a relation between the altitude on the hypotenuse in a right triangle and the two line segments it creates on the hypotenuse. It states that the geometric mean of the two segments equals the altitude.

Contents 1 Theorem and applications 2 History 3 Proof 3.1 Based on similarity 3.2 Based on the Pythagorean theorem 3.3 Based on dissection and rearrangement 3.4 Based on shear mappings 4 References 5 External links Theorem and applications Construction of √p by setting q to 1 If h denotes the altitude in a right triangle and p and q the segments on the hypotenuse then the theorem can be stated as:[1] {displaystyle h={sqrt {pq}}} or in term of areas: {displaystyle h^{2}=pq.} AM-GM inequality The latter version yields a method to square a rectangle with ruler and compass, that is to construct a square of equal area to a given rectangle. For such a rectangle with sides p and q we denote its top left vertex with D. Now we extend the segment q to its left by p (using arc AE centered on D) and draw a half circle with endpoints A and B with the new segment p+q as its diameter. Then we erect a perpendicular line to the diameter in D that intersects the half circle in C. Due to Thales' theorem C and the diameter form a right triangle with the line segment DC as its altitude, hence DC is the side of a square with the area of the rectangle. The method also allows for the construction of square roots (see constructible number), since starting with a rectangle that has a width of 1 the constructed square will have a side length that equals the square root of the rectangle's length.[1] Another application of provides a geometrical proof of the AM–GM inequality in the case of two numbers. For the numbers p and q one constructs a half circle with diameter p+q. Now the altitude represents the geometric mean and the radius the arithmetic mean of the two numbers. Since the altitude is always smaller or equal to the radius, this yields the inequality.[2] geometric mean theorem as a special case of the chord theorem: {displaystyle |CD||DE|=|AD||DB|Leftrightarrow h^{2}=pq} The theorem can also be thought of as a special case of the intersecting chords theorem for a circle, since the converse of Thales' theorem ensures that the hypotenuse of the right angled triangle is the diameter of its circumcircle.[1] The converse statement is true as well. Any triangle, in which the altitude equals the geometric mean of the two line segments created by it, is a right triangle.

History The theorem is usually attributed to Euclid (ca. 360–280 BC), who stated it as a corollary to proposition 8 in book VI of his Elements. In proposition 14 of book II Euclid gives a method for squaring a rectangle, which essentially matches the method given here. Euclid however provides a different slightly more complicated proof for the correctness of the construction rather than relying on the geometric mean theorem.[1][3] Proof Based on similarity {displaystyle triangle ABCsim triangle ADCsim triangle DBC} Proof of theorem: The triangles {displaystyle triangle ADC} and {displaystyle triangle BCD} are similar, since: consider triangles {displaystyle triangle ABC,triangle ACD} , here we have {displaystyle angle ACB=angle ADC=90^{circ }} and {displaystyle angle BAC=angle CAD} , therefore by the AA postulate {displaystyle triangle ABCsim triangle ACD} further, consider triangles {displaystyle triangle ABC,triangle BCD} , here we have {displaystyle angle ACB=angle BDC=90^{circ }} and {displaystyle angle ABC=angle CBD} , therefore by the AA postulate {displaystyle triangle ABCsim triangle BCD} Therefore, both triangles {displaystyle triangle ACD} and {displaystyle triangle BCD} are similar to {displaystyle triangle ABC} and themselves, i.e. {displaystyle triangle ACDsim triangle ABCsim triangle BCD} .

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:.[1] {displaystyle {frac {h}{p}}={frac {q}{h}},Leftrightarrow ,h^{2}=pq,Leftrightarrow ,h={sqrt {pq}}qquad (h,p,q>0)} Proof of converse: For the converse we have a triangle {displaystyle triangle ABC} in which {displaystyle h^{2}=pq} holds and need to show that the angle at C is a right angle. Now because of {displaystyle h^{2}=pq} we also have {displaystyle {tfrac {h}{p}}={tfrac {q}{h}}} . Together with {displaystyle angle ADC=angle CDB} the triangles {displaystyle triangle ADC} and {displaystyle triangle BDC} have an angle of equal size and have corresponding pairs of legs with the same ratio. This means the triangles are similar, which yields: {displaystyle angle ACB=angle ACD+angle DCB=angle ACD+(90^{circ }-angle DBC)=angle ACD+(90^{circ }-angle ACD)=90^{circ }} Based on the Pythagorean theorem Proof with the Pythagorean theorem In the setting of the geometric mean theorem there are three right triangles {displaystyle triangle ABC} , {displaystyle triangle ADC} and {displaystyle triangle DBC} , in which the Pythagorean theorem yields: {displaystyle h^{2}=a^{2}-q^{2}} , {displaystyle h^{2}=b^{2}-p^{2}} and {displaystyle c^{2}=a^{2}+b^{2}} Adding the first 2 two equations and then using the third then leads to: {displaystyle 2h^{2}=a^{2}+b^{2}-p^{2}-q^{2}=c^{2}-p^{2}-q^{2}=(p+q)^{2}-p^{2}-q^{2}=2pq} .

A division by two finally yields the formula of the geometric mean theorem.[4] Based on dissection and rearrangement Dissecting the right triangle along its altitude h yields two similar triangles, which can be augmented and arranged in two alternative ways into a larger right triangle with perpendicular sides of lengths p+h and q+h. One such arrangement requires a square of area h2 to complete it, the other a rectangle of area pq. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.

Based on shear mappings The square of the altitude can be transformed into an rectangle of equal area with sides p and q with the help of three shear mappings (shear mappings preserve the area): Shear mappings with their associated fixed lines (dotted), starting with the original square as preimage each parallelogram displays the image of a shear mapping of the figure left of it References ^ Jump up to: a b c d e *Hartmut Wellstein, Peter Kirsche: Elementargeometrie. Springer, 2009, ISBN 9783834808561, pp. 76-77 (German, online copy, p. 76, at Google Books) ^ Claudi Alsina, Roger B. Nelsen: Icons of Mathematics: An Exploration of Twenty Key Images. MAA 2011, ISBN 9780883853528, pp. 31–32 (online copy, p. 31, at Google Books) ^ Euclid: Elements, book II – prop. 14, book VI – pro6767800hshockedmake ,me uoppppp. 8, (online copy) ^ Ilka Agricola, Thomas Friedrich: Elementary Geometry. AMS 2008, ISBN 9780821843475, p. 25 (online copy, p. 25, at Google Books) External links Geometric Mean at Cut-the-Knot hide vte Ancient Greek and Hellenistic mathematics (Euclidean geometry) Mathematicians (timeline) AnaxagorasAnthemiusArchytasAristaeus the ElderAristarchusApolloniusArchimedesAutolycusBionBrysonCallippusCarpusChrysippusCleomedesCononCtesibiusDemocritusDicaearchusDioclesDiophantusDinostratusDionysodorusDomninusEratosthenesEudemusEuclidEudoxusEutociusGeminusHeliodorusHeronHipparchusHippasusHippiasHippocratesHypatiaHypsiclesIsidore of MiletusLeonMarinusMenaechmusMenelausMetrodorusNicomachusNicomedesNicotelesOenopidesPappusPerseusPhilolausPhilonPhilonidesPorphyryPosidoniusProclusPtolemyPythagorasSerenus SimpliciusSosigenesSporusThalesTheaetetusTheanoTheodorusTheodosiusTheon of AlexandriaTheon of SmyrnaThymaridasXenocratesZeno of EleaZeno of SidonZenodorus Treatises AlmagestArchimedes PalimpsestArithmeticaConics (Apollonius)CatoptricsData (Euclid)Elements (Euclid)Measurement of a CircleOn Conoids and SpheroidsOn the Sizes and Distances (Aristarchus)On Sizes and Distances (Hipparchus)On the Moving Sphere (Autolycus)Euclid's OpticsOn SpiralsOn the Sphere and CylinderOstomachionPlanisphaeriumSphaericsThe Quadrature of the ParabolaThe Sand Reckoner Problems Constructible numbers Angle trisectionDoubling the cubeSquaring the circleProblem of Apollonius Concepts and definitions Angle CentralInscribedChordCircles of Apollonius Apollonian circlesApollonian gasketCircumscribed circleCommensurabilityDiophantine equationDoctrine of proportionalityGolden ratioGreek numeralsIncircle and excircles of a triangleMethod of exhaustionParallel postulatePlatonic solidLune of HippocratesQuadratrix of HippiasRegular polygonStraightedge and compass constructionTriangle center Results In Elements Angle bisector theoremExterior angle theoremEuclidean algorithmEuclid's theoremGeometric mean theoremGreek geometric algebraHinge theoremInscribed angle theoremIntercept theoremIntersecting chords theoremIntersecting secants theoremLaw of cosinesPons asinorumPythagorean theoremTangent-secant theoremThales's theoremTheorem of the gnomon Apollonius Apollonius's theorem Other Aristarchus's inequalityCrossbar theoremHeron's formulaIrrational numbersLaw of sinesMenelaus's theoremPappus's area theoremProblem II.8 of ArithmeticaPtolemy's inequalityPtolemy's table of chordsPtolemy's theoremSpiral of Theodorus Centers CyreneLibrary of AlexandriaPlatonic Academy Other Ancient Greek astronomyGreek numeralsLatin translations of the 12th centuryNeusis construction  Ancient Greece portal •  Mathematics portal Wikimedia Commons has media related to Geometric mean theorem. Categories: AreaEuclidean plane geometryHistory of geometryTheorems about right triangles

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