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Pythagorean 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). 
In mathematics, the Pythagorean theorem—or Pythagoras' theorem—is a 
relation in Euclidean geometry among the three sides of aright triangle. It states 
that the square of the hypotenuse (the side opposite the right angle) is equal to 
the sum of the squares of the other two sides. The theorem can be written as 
an equation relating the lengths of the sides a, b and c, often called 
the Pythagorean equation:[1] 
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 (ca. 570 BC—ca. 495 BC), who by 
tradition is credited with its proof,[2][3] although it is often argued that 
knowledge of the theorem predates him. There is evidence that Babylonian 
mathematiciansunderstood the formula, although there is little surviving 
evidence that they used it in a mathematical 
framework] Mesopotamian, Indianand Chinese mathematicians have all been 
known for independently discovering the result, some even providing proofs 
of special cases. 
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.
Pythagorean proof 
The Pythagorean proof 
The Pythagorean Theorem was known long before Pythagoras, but 
he may well have been the first to prove it.[6] In any event, the 
proof attributed to him is very simple, and is called a proof by 
rearrangement. 
The two large squares shown in the figure each contain four 
identical triangles, and the only difference between the two large 
squares is that the triangles are arranged differently. Therefore, the 
white space within each of the two large squares must have equal 
area. Equating the area of the white space yields the Pythagorean 
Theorem, Q.E.D.[7] 
That Pythagoras originated this very simple proof is sometimes 
inferred from the writings of the later Greek philosopher and 
mathematicianProclus.[8] Several other proofs of this theorem are 
described below, but this is known as the Pythagorean one. 
Other forms of the theorem 
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: 
If the length of both a and b are known, then c can be calculated 
as
If the length of the hypotenuse c and of one side (a or b) are 
known, then the length of the other side can be calculated as 
or 
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 other sides, but less than their sum. 
A generalization of this theorem is the law of cosines, 
which allows the computation of the length of any side 
of any triangle, given the lengths of the other two sides 
and the angle between them. If the angle between the 
other sides is a right angle, the law of cosines reduces 
to the Pythagorean equation. 
Other proofs of the theorem 
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 Propositioncontains 370 proofs. 
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. 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 triangleABC, 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 theparallel postulate. 
Similarity of the triangles leads to the equality of ratios 
of corresponding sides: 
The first result equates the cosines of the angles θ, 
whereas the second result equates their sines. 
These ratios can be written as 
Summing these two equalities results in 
which, after simplification, expresses the 
Pythagorean theorem: 
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 
inthe Elements, and that the theory of proportions needed further 
development at that time. 
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 follow. 
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.[12] 
Illustration including the new lines 
Showing the two congruent triangles of half the area of rectangle 
BDLK and square BAGF 
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.[13] 
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. 
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 rectangle 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,[14] demonstrates that the area of the square 
on the hypotenuse is the sum of the areas of the other two 
squares.[15] This is quite distinct from the proof by similarity of 
triangles, which is conjectured to be the proof that Pythagoras used

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Pythagorean theorem

  • 1. Pythagorean 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). In mathematics, the Pythagorean theorem—or Pythagoras' theorem—is a relation in Euclidean geometry among the three sides of aright triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:[1] 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 (ca. 570 BC—ca. 495 BC), who by tradition is credited with its proof,[2][3] although it is often argued that knowledge of the theorem predates him. There is evidence that Babylonian mathematiciansunderstood the formula, although there is little surviving evidence that they used it in a mathematical framework] Mesopotamian, Indianand Chinese mathematicians have all been known for independently discovering the result, some even providing proofs of special cases. 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.
  • 2. Pythagorean proof The Pythagorean proof The Pythagorean Theorem was known long before Pythagoras, but he may well have been the first to prove it.[6] In any event, the proof attributed to him is very simple, and is called a proof by rearrangement. The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. Therefore, the white space within each of the two large squares must have equal area. Equating the area of the white space yields the Pythagorean Theorem, Q.E.D.[7] That Pythagoras originated this very simple proof is sometimes inferred from the writings of the later Greek philosopher and mathematicianProclus.[8] Several other proofs of this theorem are described below, but this is known as the Pythagorean one. Other forms of the theorem 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: If the length of both a and b are known, then c can be calculated as
  • 3. If the length of the hypotenuse c and of one side (a or b) are known, then the length of the other side can be calculated as or 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 other sides, but less than their sum. A generalization of this theorem is the law of cosines, which allows the computation of the length of any side of any triangle, given the lengths of the other two sides and the angle between them. If the angle between the other sides is a right angle, the law of cosines reduces to the Pythagorean equation. Other proofs of the theorem 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 Propositioncontains 370 proofs. Proof using similar triangles Proof using similar triangles
  • 4. 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. 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 triangleABC, 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 theparallel postulate. Similarity of the triangles leads to the equality of ratios of corresponding sides: The first result equates the cosines of the angles θ, whereas the second result equates their sines. These ratios can be written as Summing these two equalities results in which, after simplification, expresses the Pythagorean theorem: 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
  • 5. involved a theory of proportions, a topic not discussed until later inthe Elements, and that the theory of proportions needed further development at that time. 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 follow. 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).
  • 6. 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.[12] Illustration including the new lines Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF The proof is as follows:
  • 7. 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.[13] 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. 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 rectangle 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.
  • 8. This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1,[14] demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.[15] This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used