Pythagorean theorem

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

Pythagorean theorem

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