Geometric mean theorem

area of grey square = area of grey rectangle: Template:Math

In Euclidean geometry, the right triangle altitude theorem or geometric mean 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 those two segments equals the altitude.

Theorem and its converse

geometric mean theorem as a special case of the intersecting chords theorem: Template:Math Template:Math

If Template:Mvar denotes the altitude in a right triangle and Template:Mvar and Template:Mvar the segments on the hypotenuse then the theorem can be stated as: $$h=\sqrt{pq}$$ or in term of areas: $$h^2=pq.$$ 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.

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.

Applications

Construction of Template:Math by setting Template:Mvar to 1

Demonstration of AM–GM inequality

The formulation in terms of areas 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 Template:Mvar and Template:Mvar we denote its top left vertex with Template:Mvar Template:See below. Now we extend the segment Template:Mvar to its left by Template:Mvar (using arc Template:Mvar centered on Template:Mvar) and draw a half circle with endpoints Template:Mvar and Template:Mvar with the new segment Template:Math as its diameter. Then we erect a perpendicular line to the diameter in Template:Mvar that intersects the half circle in Template:Mvar. Due to Thales' theorem Template:Mvar and the diameter form a right triangle with the line segment Template:Mvar as its altitude, hence Template:Mvar 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.

Another application of this theorem provides a geometrical proof of the AM–GM inequality in the case of two numbers, since the half-circle's radius is the arithmetic mean of Template:Mvar and Template:Mvar. This radius can be drawn parallel to the geometric mean constructed as above, which shows that geometric mean is always smaller or equal to the radius, and yields the inequality.

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.

Proofs

Proof based on similarity

Template:Center

Proof of theorem

The triangles Template:Math are similar, since:

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

Therefore, both triangles Template:Math are similar to Template:Math and themselves, i.e. $$\mathrm{△}ACD\sim \mathrm{△}ABC\sim \mathrm{△}BCD.$$

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem: $$\frac{h}{p}=\frac{q}{h}⟺h^2=pq⟺h=\sqrt{pq}(h,p,q>0)$$

Proof of converse

For the converse we have a triangle Template:Math in which Template:Math holds and need to show that the angle at Template:Mvar is a right angle. Now because of Template:Math we also have $\frac{h}{p}=\frac{q}{h}.$ Together with $\mathrm{\angle }ADC=\mathrm{\angle }CDB$ the triangles Template:Math have an angle of equal size and have corresponding pairs of legs with the same ratio. This means the triangles are similar, which yields: $$\begin{array}{cc}\mathrm{\angle }ACB& =\mathrm{\angle }ACD+\mathrm{\angle }DCB\\ & =\mathrm{\angle }ACD+(90^{\circ }-\mathrm{\angle }DBC)\\ & =\mathrm{\angle }ACD+(90^{\circ }-\mathrm{\angle }ACD)\\ & =90^{\circ }\end{array}$$

Proof based on the Pythagorean theorem

Proof with the Pythagorean theorem

In the setting of the geometric mean theorem there are three right triangles Template:Math, Template:Math and Template:Math in which the Pythagorean theorem yields: $$\begin{array}{cc}{h}^2& =a^2-q^2\\ h^2& =b^2-p^2\\ c^2& =a^2+b^2\end{array}$$ Adding the first 2 two equations and then using the third then leads to: $$\begin{array}{cc}2h^2& =a^2+b^2-p^2-q^2\\ & =c^2-p^2-q^2\\ & =(p+q{)}^2-p^2-q^2\\ & =2pq\\ \therefore h^2& =pq,\end{array}$$ which finally yields the formula of the geometric mean theorem.

Proof based on dissection and rearrangement

Dissecting the right triangle along its altitude Template:Mvar yields two similar triangles, which can be augmented and arranged in two alternative ways into a larger right triangle with perpendicular sides of lengths Template:Math and Template:Math. One such arrangement requires a square of area Template:Math to complete it, the other a rectangle of area Template:Mvar. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.

Proof based on shear mappings

A square constructed on the altitude can be transformed into a rectangle of equal area with sides Template:Mvar and Template:Mvar 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

External links

• Geometric Mean at Cut-the-Knot