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 denotes the altitude in a right triangle and and denote the segments that the altitude creates on the hypotenuse, it can be stated as:
h=\sqrt{pq}
h2=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 is used in the following straightedge and compass constructions.
The
h2=pq
(The image in the Proof > Based on similarity section depicts the vertices and arches mentioned)
For a rectangle with sides and we denote its top left vertex with . Now we extend the segment to its left by (using arc centered on) and draw a half circle with endpoints and with the new segment as its diameter. Then we erect a perpendicular line to the diameter in that intersects the half circle in . As per Thales' theorem the angle between and is a right angle, the
\triangleABC
h2=pq
pq
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 is 1, since then the first version of the formula becomes
h=\sqrt{p x 1}
Another application of the theorem provides a geometrical proof of the AM–GM inequality in the case of two numbers. For the numbers and one constructs a half circle with diameter . 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]
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]
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 of theorem:
The triangles are similar, since:
Therefore, both triangles are similar to and themselves, i.e.
Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:[1]
h | = | |
p |
q | |
h |
\Leftrightarrowh2=pq\Leftrightarrowh=\sqrt{pq} (h,p,q>0)
Proof of converse:
For the converse we have a triangle in which
h2=pq
h2=pq
\tfrac{h}{p}=\tfrac{q}{h}.
\angleADC=\angleCDB
\begin{align} \angleACB&=\angleACD+\angleDCB\\ &=\angleACD+(90\circ-\angleDBC)\\ &=\angleACD+(90\circ-\angleACD)\\ &=90\circ \end{align}
\tan\alpha ⋅ \tan\beta=
h | ⋅ | |
p |
h | |
q |
\implies\tan\alpha ⋅ \cot\alpha=
h | |
pq |
\implies1=
h2 | |
pq |
\impliesh=\sqrt{pq}
In the setting of the geometric mean theorem there are three right triangles, and in which the Pythagorean theorem yields:
\begin{align} h2&=a2-q2\\ h2&=b2-p2\\ c2&=a2+b2 \end{align}
\begin{align} 2h2&=a2+b2-p2-q2\\ &=c2-p2-q2\\ &=(p+q)2-p2-q2\\ &=2pq\\ \therefore h2&=pq. \end{align}
Dissecting the right triangle along its altitude yields two similar triangles, which can be augmented and arranged in two alternative ways into a larger right triangle with perpendicular sides of lengths and . One such arrangement requires a square of area to complete it, the other a rectangle of area . Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.
The square of the altitude can be transformed into an rectangle of equal area with sides and with the help of three shear mappings (shear mappings preserve the area):