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{{Short description|Theorem about right triangles}}
[[File:Hoehensatz2.svg|thumb|upright=1.2|area of grey square = area of grey rectangle: <math> h^2=pq \Leftrightarrow h=\sqrt{pq}</math>]]

In [[Euclidean geometry]], the '''right triangle altitude theorem''' or '''geometric mean theorem''' is a relation between the [[Altitude (triangle)|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 those two segments equals the altitude.

==Theorem and its converse==
[[File:Sehnensatz hoehensatz.svg|thumb|upright=1.2|geometric mean theorem as a special case of the [[intersecting chords theorem]]:<br/><math>|CD||DE|=|AD||DB| \Leftrightarrow h^2=pq</math>]]
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:<ref name="HW"/>
:<math>h=\sqrt{pq} </math>
or in term of areas:
:<math>h^2=pq.</math>
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]].<ref name="HW"/>

== Applications ==
[[File:Root_construction_geometric_mean5.svg|thumb|upright=1.2|Construction of {{tmath|\sqrt p}} by setting {{mvar|q}} to 1]]
[[File:Am gm half circle3.svg|thumb|upright=1.2|AM-GM inequality]]
The formulation in terms of areas yields a method to square a rectangle with [[Compass and straightedge constructions|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 (geometry)|vertex]] with {{mvar|D}} (see [[#Based on similarity|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.<ref name="HW">*Hartmut Wellstein, Peter Kirsche: ''Elementargeometrie''. Springer, 2009, {{ISBN|9783834808561}}, pp. 76-77 (German, {{Google books|cFzWcl9xiGcC|online copy|page=76}}) </ref>

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.<ref>Claudi Alsina, Roger B. Nelsen: ''[[Icons of Mathematics|Icons of Mathematics: An Exploration of Twenty Key Images]]''. MAA 2011, {{ISBN|9780883853528}}, pp.&nbsp;31–32 ({{Google books|4DavMl7-aFgC|online copy|page=31}}) </ref> 

== 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 [[Euclid's elements|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.<ref name="HW"/><ref>[[Euclid]]: [[Euclid's Elements|''Elements'']],  book II – prop. 14, book VI – prop. 8, ([http://aleph0.clarku.edu/~djoyce/java/elements/elements.html online copy]) </ref>

==Proof==
===Based on similarity===
[[File:Hoehensatz.svg|thumb|upright=1.2|{{center|<math>\triangle ABC \sim \triangle ADC \sim \triangle DBC</math>}}]]
'''Proof of theorem''':

The triangles {{math|△''ADC'' , △'' BCD'' }} are [[Similar triangle#Similar triangles|similar]], since:
* consider triangles {{math|△''ABC'', △''ACD'' }}; here we have <math display=block>\angle ACB=\angle ADC=90^\circ, \quad \angle BAC=\angle CAD;</math> therefore by the [[AA postulate]] <math display=block>\triangle ABC \sim \triangle ACD .</math>
* further, consider triangles {{math|△''ABC'', △''BCD'' }}; here we have <math display=block>\angle ACB=\angle BDC= 90^\circ, \quad \angle ABC=\angle CBD;</math> therefore by the AA postulate <math display=block>\triangle ABC \sim \triangle BCD.</math>

Therefore, both triangles {{math|△''ACD'', △''BCD''}} are similar to {{math|△''ABC''}} and themselves, i.e. <math display=block>\triangle ACD \sim \triangle ABC \sim \triangle BCD.</math>

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:<ref name="HW"/>
:<math> \frac{h}{p}=\frac{q}{h}\,\Leftrightarrow\,h^2=pq\,\Leftrightarrow\,h=\sqrt{pq}\qquad (h,p,q> 0)</math>

'''Proof of converse:'''

For the converse we have a triangle {{math|△''ABC''}} in which <math>h^2=pq</math> holds and need to show that the angle at {{mvar|C}} is a right angle. Now because of <math>h^2=pq</math> we also have <math>\tfrac{h}{p}=\tfrac{q}{h}.</math> Together with <math>\angle ADC=\angle CDB </math> 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:
:<math>\begin{align}
\angle ACB &= \angle ACD +\angle DCB \\
           &= \angle ACD+(90^\circ-\angle DBC) \\
           &= \angle ACD+(90^\circ-\angle ACD) \\
           &= 90^\circ
\end{align}</math>

===Based on the Pythagorean theorem===
[[File:Hoehensatz beweis pythagoras.svg|thumb|upright=1.0|Proof with the Pythagorean theorem]]
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:
:<math>\begin{align}
h^2 &= a^2-q^2 \\
h^2 &= b^2-p^2 \\
c^2 &= a^2+b^2
\end{align}</math>
Adding the first 2 two equations and then using the third then leads to:
:<math>\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}</math>
which finally yields the formula of the geometric mean theorem.<ref>[[Ilka Agricola]], Thomas Friedrich: ''Elementary Geometry''. AMS 2008, {{ISBN|9780821843475}}, p.&nbsp;25  ({{Google books|LLXxBwAAQBAJ|online copy|page=25}}) </ref>

===Based on dissection and rearrangement===
[[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|''h''<sup>2</sup>}} 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.

===Based on shear mappings===
The square of the altitude can be transformed into an rectangle of equal area with sides {{mvar|p}} and {{mvar|q}} with the help of three [[shear mapping]]s (shear mappings preserve the area):

[[File:Scherungen alle2.svg|thumb|center|upright=4.0|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==
<references/>

== External links ==
*[https://www.cut-the-knot.org/pythagoras/GeometricMean.shtml ''Geometric Mean''] at [[Cut-the-Knot]]

{{Ancient Greek mathematics}}

{{commonscat|Geometric mean theorem}}

[[Category:Area]]
[[Category:Articles containing proofs]]
[[Category:Euclidean plane geometry]]
[[Category:History of geometry]]
[[Category:Theorems about right triangles]]