————————————————- Pythagorean Theorem 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] here c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. These two formulations show two fundamental aspects of this theorem: it is both a statement about areas and about lengths. Tobias Dantzig refers to these as areal and metric interpretations. [2][3] Some proofs of the theorem are based on one interpretation, some upon the other. Thus, Pythagoras’ theorem stands with one foot in geometry and the other in algebra, a connection made clear originally byDescartes in his work La Geometrie, and extending today into other branches of mathematics. 4] The Pythagorean theorem has been modified to apply outside its original domain. A number of these generalizations are described below, including extension to many-dimensional Euclidean 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 is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof,[5][6] although it is often argued that knowledge of the theorem predates him. There is much evidence that Babylonian mathematicians understood the formula, although there is little surviving evidence that they fitted it into a mathematical framework. [7]) “[To the Egyptians and Babylonians] mathematics provided practical tools in the form of “recipes” designed for specific calculations. Pythagoras, on the other hand, was one of the first to grasp numbers as abstract entities that exist in their own right. ”[8] In addition to a separate section devoted to the history of Pythagoras’ theorem, historical asides and sources are found in many of the other subsections.

The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power. The article ends with a section on pop references to the theorem. 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). ————————————————- 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, Pythagoras’ theorem can be expressed as the Pythagorean equation: r, solved for c: If c is known, and the length of one of the legs must be found, the following equations can be used: or The Pythagorean equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. 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. [9] [edit]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 ? 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: The first result equates the cosine of each angle ? and the second result equates the sines. These ratios can be written as: Summing these two equalities, we obtain which, tidying up, is the Pythagorean theorem: This is a metric proof in the sense of Dantzig, one that depends on lengths, not areas.

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. [10][11] 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: 1. 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. 3. The area of a rectangle is equal to the product of two adjacent sides. 4. 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. [12] Illustration including the new lines The proof is as follows: 1. Let ACB be a right-angled triangle with right angle CAB. 2.

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. [13] 3. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively. 4. Join CF and AD, to form the triangles BCF and BDA. 5. 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 1.

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. 2. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be congruent to triangle FBC. 3. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD. 4. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC. 5. Therefore rectangle BDLK must have the same area as square BAGF = AB2. 6. 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 8. Since BD = KL, BD* BK + KL ? KC = BD(BK + KC) = BD ? BC 9. Therefore AB2 + AC2 = BC2, since CBDE is a square. lgebraic proof Outer square with identical right triangles of adjacent sides a and b inserted, forming an interior square of side c. An algebraic proof is provided by the following reasoning. A point is marked on each side of the outer square, dividing each side into two segments of lengths a and b. As shown in the illustration, the result is a large square with identical right triangles in its corners.

Each pair of triangles from opposite corners forms a rectangle with area ab, so the area of all four triangles is: * Area of 4 triangles = 2ab. The center figure has equal sides because all the four triangles forming its border are the same. However, it might be any kind of rhombus. Next, the center figure is shown to be a square: The a-side angle and b-side angle of each of these triangles are complementary angles, that is, they sum to a right angle. However, the corner angle c of the inner figure and the two side-angles a and b sum to a straight angle.

So each of the angles of the blue area in the middle is a right angle, making this area a square with side length c. The area of this square is c2. Thus the area of everything together is given by: * Area of complete figure = c2 + 2ab. However, as the large square has sides of length a + b, we can also calculate its area as: * Area of complete figure = (a + b)2 = a2 + 2ab + b2. Equating the two expressions for the total area, the result is: A term of 2ab can be subtracted from both sides of the equation, leaving: which is Pythagoras’ theorem.