Geometric mean theorem

area of grey square = area of grey rectangle: ${\displaystyle h^{2}=pq\Leftrightarrow h={\sqrt {pq}}}$

In Euclidean geometry, the geometric mean theorem or right triangle altitude theorem is 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.

Expressed as a mathematical formula, if h denotes the altitude in a right triangle and p and q denote the segments that the altitude creates on the hypotenuse, it can be stated as: ^[1]

${\displaystyle h={\sqrt {pq}}}$

or in term of areas:

${\displaystyle h^{2}=pq}$.

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.

Applications

Straightedge and compass construction

The theorem is used in the following straightedge and compass constructions.

Squaring a rectangle

The ${\displaystyle h^{2}=pq}$ version of the formula yields a method to construct a square of equal area to a given rectangle through the following steps:

(The image in the Proof > Based on similarity section depicts the vertices and arches mentioned)
For 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. As per Thales' theorem the angle between AC and CB is a right angle, the ${\displaystyle \triangle ABC}$ is a right triangle, and so the theorem applies: its identity ${\displaystyle h^{2}=pq}$ directly shows that a square with the area of the rectangle (equal to ${\displaystyle pq}$) can be drawn by using exactly DC as the squares' side, because DC is the h.

Constructing a square root

Construction of ⁠${\displaystyle {\sqrt {p}}}$⁠ by setting q to 1 (note some letters are different than used throughout the rest of the article, notably the articles' DC is BF on the image)

The above method from Squaring a rectangle also allows for the construction of square roots (see constructible number) by starting with a rectangle whose side q is 1, since then the first version of the formula becomes ${\displaystyle h={\sqrt {p\times 1}}}$, showing that DC (h in the formula) will readily be the root of p. ^[1]

Relation with other theorems

AM-GM inequality

Another application of the theorem 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]

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

${\displaystyle \triangle ABC\sim \triangle ADC\sim \triangle DBC}$

Proof of theorem:

The triangles △ADC , △ BCD are similar, since:

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

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

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 △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 △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:

${\displaystyle {\begin{aligned}\angle ACB&=\angle ACD+\angle DCB\\&=\angle ACD+(90^{\circ }-\angle DBC)\\&=\angle ACD+(90^{\circ }-\angle ACD)\\&=90^{\circ }\end{aligned}}}$

Based on trigonometric ratio

${\displaystyle \tan \alpha \cdot \tan \beta ={\frac {h}{p}}\cdot {\frac {h}{q}}\,\implies \tan \alpha \cdot \cot \alpha ={\frac {h}{pq}}\implies 1={\frac {h^{2}}{pq}}\implies h={\sqrt {pq}}}$

Based on the Pythagorean theorem

Proof with the Pythagorean theorem

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

${\displaystyle {\begin{aligned}h^{2}&=a^{2}-q^{2}\\h^{2}&=b^{2}-p^{2}\\c^{2}&=a^{2}+b^{2}\end{aligned}}}$

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

${\displaystyle {\begin{aligned}2h^{2}&=a^{2}+b^{2}-p^{2}-q^{2}\\&=c^{2}-p^{2}-q^{2}\\&=(p+q)^{2}-p^{2}-q^{2}\\&=2pq\\\therefore \ h^{2}&=pq.\end{aligned}}}$

which 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

1. • Hartmut Wellstein, Peter Kirsche: Elementargeometrie. Springer, 2009, ISBN   9783834808561, pp. 76-77 (German, online copy , p. 76, at Google Books)
2. 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)
3. Euclid: Elements, book II – prop. 14, book VI – pro6767800hshockedmake ,me uoppppp. 8, (online copy)
4. 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