# Geometric mean theorem

(Redirected from Right triangle altitude theorem)

The right triangle altitude theorem or geometric mean theorem is a result in elementary geometry that describes a relation between the lengths of 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

area of grey square = area of grey rectangle
${\displaystyle h^{2}=pq\Leftrightarrow h={\sqrt {pq}}}$
geometric mean theorem as a special case of the chord theorem:
${\displaystyle |CE||EF|=|AE||EB|\Leftrightarrow h^{2}=pq}$

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:

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

or in term of areas:

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

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.

The geometric mean theorem can also be thought of a special case of the intersecting chords theorem for a circle, since converse of Thales' theorem ensures that the hypothenuse of the right angled triangle is the diameter of its circumcircle.

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. In proposition 14 of book II Euclid gives a method for squaring a rectangle, which essentially matches the method given here. Euclid however provides different slightly more complicated proof for the correctness of the construction rather than relying on the geometric mean theorem.

## Proof

### Based on similarity

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 ${\displaystyle \triangle ABC\sim \triangle ADC}$
• 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 ${\displaystyle \triangle ABC\sim \triangle BCD}$

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

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

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

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:
${\displaystyle \angle ACB=\angle ACD+\angle DCB=\angle ACD+(90^{\circ }-\angle ACD)=90^{\circ }}$

### 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.