Geometric mean theorem

File:Sehnensatz hoehensatz.svg:
$|CD||DE|=|AD||DB|\; \backslash Leftrightarrow\; h^2=pq$]]

If {{mvar|h}} denotes the altitude in a right triangle and {{mvar|p}} and {{mvar|q}} the segments on the hypotenuse then the theorem can be stated as:

:$h=\backslash sqrt\{pq\}$

or in term of areas:

:$h^2=pq.$

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.

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.

File:Root_construction_geometric_mean5.svg

File:Am gm half circle3.svg

The formulation in terms of areas 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 {{mvar|p}} and {{mvar|q}} we denote its top left vertex with {{mvar|D}} (see the Proof > Based on similarity section for a graphic of the construction). Now we extend the segment {{mvar|q}} to its left by {{mvar|p}} (using arc {{mvar|{{overline|AE}}}} centered on {{mvar|D}}) and draw a half circle with endpoints {{mvar|A}} and {{mvar|B}} with the new segment {{math|p + q}} as its diameter. Then we erect a perpendicular line to the diameter in {{mvar|D}} that intersects the half circle in {{mvar|C}}. Due to Thales' theorem {{mvar|C}} and the diameter form a right triangle with the line segment {{mvar|{{overline|DC}}}} as its altitude, hence {{mvar|{{overline|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.*Hartmut Wellstein, Peter Kirsche: Elementargeometrie. Springer, 2009, {{ISBN|9783834808561}}, pp. 76-77 (German, {{Google books|cFzWcl9xiGcC|online copy|page=76}})

Another application of this theorem provides a geometrical proof of the AM–GM inequality in the case of two numbers. For the numbers {{mvar|p}} and {{mvar|q}} one constructs a half circle with diameter {{math|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.Claudi Alsina, Roger B. Nelsen: Icons of Mathematics: An Exploration of Twenty Key Images. MAA 2011, {{ISBN|9780883853528}}, pp. 31–32 ({{Google books|4DavMl7-aFgC|online copy|page=31}})

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.Euclid: Elements, book II – prop. 14, book VI – prop. 8, ([http://aleph0.clarku.edu/~djoyce/java/elements/elements.html online copy])

File:Hoehensatz.svg

Proof of theorem:

The triangles {{math|△ADC , △ BCD }} are similar, since:

• consider triangles {{math|△ABC, △ACD }}; here we have $$\backslash angle\; ACB=\backslash angle\; ADC=90^\backslash circ,\; \backslash quad\; \backslash angle\; BAC=\backslash angle\; CAD;$$ therefore by the AA postulate $$\backslash triangle\; ABC\; \backslash sim\; \backslash triangle\; ACD\; .$$
• further, consider triangles {{math|△ABC, △BCD }}; here we have $$\backslash angle\; ACB=\backslash angle\; BDC=\; 90^\backslash circ,\; \backslash quad\; \backslash angle\; ABC=\backslash angle\; CBD;$$ therefore by the AA postulate $$\backslash triangle\; ABC\; \backslash sim\; \backslash triangle\; BCD.$$

Therefore, both triangles {{math|△ACD, △BCD}} are similar to {{math|△ABC}} and themselves, i.e. $$\backslash triangle\; ACD\; \backslash sim\; \backslash triangle\; ABC\; \backslash sim\; \backslash triangle\; BCD.$$

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:

:$\backslash frac\{h\}\{p\}=\backslash frac\{q\}\{h\}\backslash ,\backslash Leftrightarrow\backslash ,h^2=pq\backslash ,\backslash Leftrightarrow\backslash ,h=\backslash sqrt\{pq\}\backslash qquad\; (h,p,q>\; 0)$

Proof of converse:

For the converse we have a triangle {{math|△ABC}} in which $h^2=pq$ holds and need to show that the angle at {{mvar|C}} is a right angle. Now because of $h^2=pq$ we also have $\backslash tfrac\{h\}\{p\}=\backslash tfrac\{q\}\{h\}.$ Together with $\backslash angle\; ADC=\backslash angle\; CDB$ the triangles {{math|△ADC, △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:

:$\backslash begin\{align\}$

\angle ACB &= \angle ACD +\angle DCB \\

&= \angle ACD+(90^\circ-\angle DBC) \\

&= \angle ACD+(90^\circ-\angle ACD) \\

&= 90^\circ

\end{align}

File:Hoehensatz beweis pythagoras.svg

In the setting of the geometric mean theorem there are three right triangles {{math|△ABC}}, {{math|△ADC}} and {{math|△DBC}} in which the Pythagorean theorem yields:

:$\backslash begin\{align\}$

h^2 &= a^2-q^2 \\

h^2 &= b^2-p^2 \\

c^2 &= a^2+b^2

\end{align}

Adding the first 2 two equations and then using the third then leads to:

:$\backslash begin\{align\}$

2h^2 &= a^2+b^2-p^2-q^2 \\

&= c^2-p^2-q^2 \\

&= (p+q)^2-p^2-q^2 \\

&= 2pq \\

\therefore \ h^2 &= pq.

\end{align}

which finally yields the formula of the geometric mean theorem.Ilka Agricola, Thomas Friedrich: Elementary Geometry. AMS 2008, {{ISBN|9780821843475}}, p. 25 ({{Google books|LLXxBwAAQBAJ|online copy|page=25}})

File:Geometrischer Höhensatzbeweis.svg

Dissecting the right triangle along its altitude {{mvar|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 {{math|p + h}} and {{math|q + h}}. One such arrangement requires a square of area {{math|h²}} to complete it, the other a rectangle of area {{mvar|pq}}. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.

A square constructed on the altitude can be transformed into a rectangle of equal area with sides {{mvar|p}} and {{mvar|q}} with the help of three shear mappings (shear mappings preserve the area):

File:Scherungen alle2.svg

• [https://www.cut-the-knot.org/pythagoras/GeometricMean.shtml Geometric Mean] at Cut-the-Knot

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{{commonscat|Geometric mean theorem}}

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