Pythagorean theorem

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The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

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In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle (right-angled triangle). In terms of areas, it states:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:^[1]

$a^2 + b^2 = c^2\!\,$

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof,^[2]^[3] although it is often argued that knowledge of the theorem predates him. There is evidence that Babylonian mathematicians understood the formula, although there is little surviving evidence that they fitted it into a mathematical framework.^[4]^[5]

The theorem has numerous proofs, possibly the most of any mathematical theorem. These are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. The theorem can be generalized in various ways, including higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but n-dimensional solids. The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound.

[ Other forms

As pointed out in the introduction, if c denotes the length of the hypotenuse and a and b denote the lengths of the other two sides, the Pythagorean theorem can be expressed as the Pythagorean equation:

$a^2 + b^2 = c^2\,$

If the length of both a and b are known, then c can be calculated as follows:

$c = \sqrt{a^2 + b^2}. \,$

If the length of hypotenuse c and one leg (a or b) are known, then the length of the other leg can be calculated with the following equations:

$a = \sqrt{c^2 - b^2}. \,$

$b = \sqrt{c^2 - a^2}. \,$

The Pythagorean equation relates the sides of a right triangle in a simple way, so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the legs, but less than the sum of them.

A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle, the law of cosines reduces to the Pythagorean equation.

[ Proofs

This theorem may have more known proofs than any other (the law of quadratic reciprocity being another contender for that distinction); the book The Pythagorean Proposition contains 370 proofs.^[6]

[ Proof using similar triangles

Proof using similar triangles

This proof is based on the proportionality of the sides of two similar triangles, that is, upon the fact that the ratio of any two corresponding sides of similar triangles is the same regardless of the size of the triangles.

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. Point H divides the length of the hypotenuse c into parts d and e. The new triangle ACH is similar to triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well, marked as θ in the figure. By a similar reasoning, the triangle CBH is also similar to ABC. The proof of similarity of the triangles requires the Triangle postulate: the sum of the angles in a triangle is two right angles, and is equivalent to the parallel postulate. Similarity of the triangles leads to the equality of ratios of corresponding sides:

$\frac{a}{c}=\frac{e}{a} \text{ and } \frac{b}{c}=\frac{d}{b}.\,$

The first result equates the cosine of each angle θ and the second result equates the sines.

These ratios can be written as:

$a^2=c\times e \text{ and }b^2=c\times d. \,$

Summing these two equalities, we obtain

$a^2+b^2=c\times e+c\times d=c\times(d+e)=c^2 ,\,\!$

which, tidying up, is the Pythagorean theorem:

$a^2+b^2=c^2 \ .\,\!$

The role of this proof in history is the subject of much speculation. The underlying question is why Euclid did not use this proof, but invented another. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the Elements, and that the theory of proportions needed further development at that time.^[7]^[8]

[ Euclid's proof

Proof in Euclid's Elements

In outline, here is how the proof in Euclid's Elements proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details are next.

Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.

For the formal proof, we require four elementary lemmata:

If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side).
The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
The area of a rectangle is equal to the product of two adjacent sides.
The area of a square is equal to the product of two of its sides (follows from 3).

Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.^[9]

Illustration including the new lines

The proof is as follows:

Let ACB be a right-angled triangle with right angle CAB.
On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.^[10]
From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
Join CF and AD, to form the triangles BCF and BDA.
Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.

Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF
Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
Since AB is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.
Since A-K-L is a straight line, parallel to BD, then parallelogram BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2)
Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
Therefore rectangle BDLK must have the same area as square BAGF = AB².
Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC².
Adding these two results, AB² + AC² = BD × BK + KL × KC
Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
Therefore AB² + AC² = BC², since CBDE is a square.

This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1,^[11] demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.^[12] This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.^[8]^[13]

[ Proof by rearrangement

The leftmost animation consists of a large square, side a + b, containing four identical right triangles. The triangles are shown in two arrangements, the first of which leaves a square c² uncovered, the second of which leaves two squares a² and b² uncovered. But as these uncovered areas with the areas of the four triangles fill the large square, the uncovered areas must be equal, so a² + b² = c².

A second proof is given by the middle animation. A large square is formed with area c², from four identical right triangles with sides a, b and c, fitted around a small central square. Then two rectangles are formed with sides a and b by moving the triangles. Combining the smaller square with these rectangles produces two squares of areas a² and b², which must have the same area as the initial large square.[14]

The third, rightmost image also gives a proof. The upper two squares are divided as shown by the blue and green shading, into pieces that when rearranged can be made to fit in the lower square on the hypotenuse – or conversely the large square can be divided as shown into pieces that fill the other two. This shows the area of the large square equals that of the two smaller ones.^[15]

Proof by rearrangement of four identical right triangles

Animation showing another proof by rearrangement

Proof using an elaborate rearrangement

[ Algebraic proofs

Diagram of the two algebraic proofs.

The theorem can be proved algebraically using four copies of a right triangle with sides a, b and c, arranged inside a square with side c as in the top half of the diagram.^[16] The triangles are similar with area $\tfrac12ab$ , while the small square has side b − a and area (b − a)². The area of the large square is therefore

$(b-a)^2+4\frac{ab}{2} = (b-a)^2+2ab = a^2+b^2. \,$

But this is a square with side c and area c², so

$c^2 = a^2 + b^2. \,$

A similar proof uses four copies of the same triangle arranged symmetrically around a square with side c, as shown in the lower part of the diagram.^[17] This results in a larger square, with side a + b and area (a + b)². The four triangles and the square side c must have the same area as the larger square,

$(b+a)^2 = c^2 + 4\frac{ab}{2} = c^2+2ab,\,$

giving

$c^2 = (b+a)^2 - 2ab = a^2 + b^2.\,$

Diagram of Garfield's proof.

A related proof was published by future U.S. President James A. Garfield.^[18]^[19] Instead of a square it uses a trapezoid, which can be constructed from the square in the second of the above proofs by bisecting along a diagonal of the inner square, to give the trapezoid as shown in the diagram. The area of the trapezoid can be calculated to be half the area of the square, that is

$\frac{1}{2}(b+a)^2.$

The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of $\frac{1}{2}$ , which is removed by multiplying by two to give the result.

[ Proof using differentials

One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse and employing calculus.^[20]^[21]^[22]

The triangle ABC is a right triangle, as shown in the upper part of the diagram, with BC the hypotenuse. At the same time the triangle lengths are measured as shown, with the hypotenuse of length y, the side AC of length x and the side AB of length a, as seen in the lower diagram part.

Diagram for differential proof

If x is increased by a small amount dx by extending the side AC slightly to D, then y also increases by dy. These form two sides of a triangle, CDE, which (with E chosen so CE is perpendicular to the hypotenuse) is a right triangle approximately similar to ABC. Therefore the ratios of their sides must be the same, that is:

$\frac{dy}{dx} = \frac xy.$

This can be rewritten as follows:

$y \cdot dy - x \cdot dx = 0.\,$

This is a differential equation which is solved to give

$y^2 - x^2 = C,\,$

And the constant can be deduced from x = 0, y = a to give the equation

$y^2 = x^2 + a^2\,$

This is more of an intuitive proof than a formal one: it can be made more rigorous if proper limits are used in place of dx and dy.

[ Converse

The converse of the theorem is also true:^[23]

For any three positive numbers a, b, and c such that a² + b² = c², there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.

An alternative statement is:

For any triangle with sides a, b, c, if a² + b² = c², then the angle between a and b measures 90°.

This converse also appears in Euclid's Elements (Book I, Proposition 48):^[24]

"If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right."

It can be proven using the law of cosines or as follows:

Let ABC be a triangle with side lengths a, b, and c, with a² + b² = c². Construct a second triangle with sides of length a and b containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length c = √a² + b², the same as the hypotenuse of the first triangle. Since both triangles' sides are the same lengths a, b and c, the triangles are congruent and must have the same angles. Therefore, the angle between the side of lengths a and b in the original triangle is a right angle.

The above proof of the converse makes use of the Pythagorean Theorem itself. The converse can also be proven without assuming the Pythagorean Theorem.^[25]^[26]

A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Let c be chosen to be the longest of the three sides and a + b > c (otherwise there is no triangle according to the triangle inequality). The following statements apply:^[27]

If a² + b² = c², then the triangle is right.
If a² + b² > c², then the triangle is acute.
If a² + b² < c², then the triangle is obtuse.

Edsger Dijkstra has stated this proposition about acute, right, and obtuse triangles in this language:

sgn(α + β − γ) = sgn(a² + b² − c²),

where α is the angle opposite to side a, β is the angle opposite to side b, γ is the angle opposite to side c, and sgn is the sign function.^[28]

[ Consequences and uses of the theorem

[ Pythagorean triples

Main article: Pythagorean triple

A Pythagorean triple has three positive integers a, b, and c, such that a² + b² = c². In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths.^[1] Evidence from megalithic monuments in Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c). Some well-known examples are (3, 4, 5) and (5, 12, 13).

A primitive Pythagorean triple is one in which a, b and c are coprime (the greatest common divisor of a, b and c is 1).

The following is a list of primitive Pythagorean triples with values less than 100:

(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)

[ Incommensurable lengths

The spiral of Theodorus: A construction for line segments with lengths whose ratios are the square root of a positive integer

One of the consequences of the Pythagorean theorem is that line segments whose lengths are incommensurable (so the ratio of which is not a rational number) can be constructed using a straightedge and compass. Pythagoras' theorem enables construction of incommensurable lengths because the hypotenuse of a triangle is related to the sides by the square root operation.

The figure on the right shows how to construct line segments whose lengths are in the ratio of the square root of any positive integer.^[29] Each triangle has a side (labeled "1") that is the chosen unit for measurement. In each right triangle, Pythagoras' theorem establishes the length of the hypotenuse in terms of this unit. If a hypotenuse is related to the unit by the square root of a positive integer that is not a perfect square, it is a realization of a length incommensurable with the unit, such as √2, √3, √5 . For more detail, see Quadratic irrational.

Incommensurable lengths conflicted with the Pythagorean school's concept of numbers as only whole numbers. The Pythagorean school dealt with proportions by comparison of integer multiples of a common subunit.^[30] According to one legend, Hippasus of Metapontum (ca. 470 B.C.) was drowned at sea for making known the existence of the irrational or incommensurable.^[31]^[32]

[ Complex numbers

The absolute value of a complex number z is the distance r from z to the origin

For any complex number

$z = x + iy,\,$

the absolute value or modulus is given by

$r = |z|=\sqrt{x^2 + y^2}.\,$

So the three quantities, r, x and y are related by the Pythagorean equation,

$r^2 = x^2 + y^2.\,$

Note that r is defined to be a positive number or zero but x and y can be negative as well as positive. Geometrically r is the distance of the z from zero or the origin O in the complex plane.

This can be generalised to find the distance between two points, z₁ and z₂ say. The required distance is given by

$|z_1 - z_2|=\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2},\,$

so again they are related by a version of the Pythagorean equation,

$|z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2.\,$

[ Euclidean distance in various coordinate systems

The distance formula in Cartesian coordinates is derived from the Pythagorean theorem.^[33] If (x₁, y₁) and (x₂, y₂) are points in the plane, then the distance between them, also called the Euclidean distance, is given by

$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.$

More generally, in Euclidean n-space, the Euclidean distance between two points, $A\,=\,(a_1,a_2,\dots,a_n)$ and $B\,=\,(b_1,b_2,\dots,b_n)$ , is defined, by generalization of the Pythagorean theorem, as:

$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} = \sqrt{\sum_{i=1}^n (a_i-b_i)^2}.$

If Cartesian coordinates are not used, for example, if polar coordinates are used in two dimensions or, in more general terms, if curvilinear coordinates are used, the formulas expressing the Euclidean distance are more complicated than the Pythagorean theorem, but can be derived from it. A typical example where the straight-line distance between two points is converted to curvilinear coordinates can be found in the applications of Legendre polynomials in physics. The formulas can be discovered by using Pythagoras' theorem with the equations relating the curvilinear coordinates to Cartesian coordinates. For example, the polar coordinates (r, θ) can be introduced as:

$x = r \cos \theta, \ y = r \sin \theta.\,$

Then two points with locations (r₁, θ₁) and (r₂, θ₂) are separated by a distance s:

$s^2 = (x_1 - x_2)^2 + (y_1-y_2)^2 = (r_1 \cos \theta_1 -r_2 \cos \theta_2 )^2 + (r_1 \sin \theta_1 -r_2 \sin \theta_2)^2.\,$

Performing the squares and combining terms, the Pythagorean formula for distance in Cartesian coordinates produces the separation in polar coordinates as:

$\begin{align}s^2 &= r_1^2 +r_2^2 -2 r_1 r_2 \left( \cos \theta_1 \cos \theta_2 +\sin \theta_1 \sin \theta_2 \right)\\ &= r_1^2 +r_2^2 -2 r_1 r_2 \cos \left( \theta_1 - \theta_2\right)\\ &=r_1^2 +r_2^2 -2 r_1 r_2 \cos \Delta \theta \end{align} \ ,$

using the trigonometric product-to-sum formulas. This formula is the law of cosines, sometimes called the Generalized Pythagorean Theorem.^[34] From this result, for the case where the radii to the two locations are at right angles, the enclosed angle Δθ = π/2, and the form corresponding to Pythagoras' theorem is regained: $s^2 = r_1^2 + r_2^2.\,$ The Pythagorean theorem, valid for right triangles, therefore is a special case of the more general law of cosines, valid for arbitrary triangles.

[ Pythagorean trigonometric identity

Main article: Pythagorean trigonometric identity

Similar right triangles showing sine and cosine of angle θ

In a right triangle with sides a, b and hypotenuse c, trigonometry determines the sine and cosine of the angle θ between side a and the hypotenuse as:

$\sin \theta = \frac{b}{c}, \quad \cos \theta = \frac{a}{c}.$

From that it follows:

${\cos}^2 \theta + {\sin}^2 \theta = \frac{a^2 + b^2}{c^2} = 1,$

where the last step applies Pythagoras' theorem. This relation between sine and cosine sometimes is called the fundamental Pythagorean trigonometric identity.^[35] In similar triangles, the ratios of the sides are the same regardless of the size of the triangles, and depend upon the angles. Consequently, in the figure, the triangle with hypotenuse of unit size has opposite side of size sin θ and adjacent side of size cos θ in units of the hypotenuse.

[ Relation to the cross product

The area of a parallelogram as a cross product; vectors a and b identify a plane and a × b is normal to this plane.

The Pythagorean theorem relates the cross product and dot product in a similar way:^[36]

$\|\mathbf{a} \times \mathbf{b}\|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = \|\mathbf{a}\|^2 \|\mathbf{b}\|^2\,$

This can be seen from the definitions of the cross product and dot product, as

$\begin{align} \mathbf{a} \times \mathbf{b} &= ab \mathbf{n} \sin{\theta} \\ \mathbf{a} \cdot \mathbf{b} &= ab \cos{\theta}\end{align}$

with n a unit vector normal to both a and b. The relationship follows from these definitions and the Pythagorean trigonometric identity.

This can also be used to define the cross product. By rearranging the following equation is obtained

$\|\mathbf{a} \times \mathbf{b}\|^2 = \|\mathbf{a}\|^2 \|\mathbf{b}\|^2 - (\mathbf{a} \cdot \mathbf{b})^2\,$

This can be considered as a condition on the cross product and so part of its definition, for example in seven dimensions.^[37]^[38]

[ Generalizations

[ Similar figures on the three sides

The Pythagorean theorem was generalized by Euclid in his Elements to extend beyond the areas of squares on the three sides to similar figures:^[39]

If one erects similar figures (see Euclidean geometry) on the sides of a right triangle, then the sum of the areas of the two smaller ones equals the area of the larger one.

The basic idea behind this generalization is that the area of a plane figure is proportional to the square of any linear dimension, and in particular is proportional to the square of thee length of any side. Thus, if similar figures with areas A, B and C are erected on sides with lengths a, b and c then:

Source: this wikipedia article, under CC-BY-SA.

Pythagorean theorem

Pythagorean theorem

Contents

[ Other forms

[ Proofs

[ Proof using similar triangles

[ Euclid's proof

[ Proof by rearrangement

[ Algebraic proofs

[ Proof using differentials

[ Converse

[ Consequences and uses of the theorem

[ Pythagorean triples

[ Incommensurable lengths

[ Complex numbers

[ Euclidean distance in various coordinate systems

[ Pythagorean trigonometric identity

[ Relation to the cross product

[ Generalizations

[ Similar figures on the three sides

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