Geometric mean theorem

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.

Theorem and application

File:Hoehensatz.svg

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

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:

$h=\sqrt{pq}$

or in term of areas:

$h^2=pq.$

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 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.

Historically the theorem is attributed to Euclid (ca. 360–280 BC), who stated it as a corollary to proposition 8 in book VI of his Elements and used it in proposition 14 of book II to square a rectangle.

Proof

Based on similarity

Proof of theorem:

The triangles $\triangle ADC$ and $\triangle BDC$ are similar, since with $\angle ADC=\angle CDB$ and $\angle CAD=90^\circ -\angle DBC=\angle BCD$ they have two angles of equal size. The similarity yields the following equation:

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

Proof of converse:

For the converse we have a triangle $\triangle ABC$ in which $h^2=pq$ holds and need to show that the angle at C is a right angle. Now because of $h^2=pq$ we also have $\tfrac{h}{p}=\tfrac{q}{h}$. Together with $\angle ADC=\angle CDB$ the triangles $\triangle ADC$ and $\triangle BDC$ have an angle of equal size and have corresponding pairs of legs with the same ratio. This means the triangles are similar, which yields:

$\angle ACB=\angle ACD +\angle DCB=\angle ACD+(90^\circ-\angle ACD)=90^\circ$

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.

References

• 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)
• Ilka Agricola, Thomas Friedrich: Elementary Geometry. AMS 2008, ISBN 9780821843475, p. 25 (online copy, p. 25, at Google Books)
• Euclid: Elements, book II – prop. 14, book VI – prop. 8, (online copy)