Mouni asked in Science & MathematicsMathematics · 1 decade ago

# Can any one help me with a proof of Pythagoras theorem?

Please note not to send similar triangles and activity based proof for this theorem.

Relevance
• 1 decade ago

trfe are a lot of proof in the link

• 1 decade ago

Go look it up in any elementary geometry book,

or in wikipedia.org/wiki/Pythagorean_theorem.

Euclid's proof is too long and clumsy to give here.

The ancient Hindu proof is brilliant - just 2 pictures (one with 4 identical triangles at the corners srrounding square of the longest side. second picture rearranges them in 2 pairs on each side, leaving the squares of the adjacnet sides). The proof is then one word: "Look!"

• Anonymous
4 years ago

Draw a right triangle ABC.Draw a perpendicular to the hypotenuse from the opposite vertex and then proof the theorem by similarity. I can't give u the proof as I can't sketch the diagram but u can find it in any 10th standard book.

• 1 decade ago

imagine a triangle with squares on each side. the hypotenuse is c.

so the area of each square if the sides of the triangle is a,b,c is a2,b2,c2 respectively. So, the area of the whole thing is a2 + b2 + c2 + ab/2

and therefore a2 + b2 = c2

• 1 decade ago

A^2+B^2=C^2 ...

What more "Proof" do you need?

The link should be able to answer any questions I have not...

• Anonymous

opposite squared plus adjacent square = hypotenuse squared

opposite = y

hypotenuse = c squared

• 1 decade ago

Proofs

This is a theorem that may have more known proofs than any other (the law of quadratic reciprocity being also a contender for that distinction); the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs.

Some arguments based on trigonometric identities (such as Taylor series for sine and cosine) have been proposed as proofs for the theorem. However, since all the fundamental trigonometric identities are proved using the Pythagorean theorem, there cannot be any trigonometric proof. (See also begging the question.)

 Proof using similar triangles

Proof using similar triangles.Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar 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. The new triangle ACH is similar to our 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. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As

so

These can be written as

Summing these two equalities, we obtain

In other words, the Pythagorean theorem:

 Euclid's proof

Proof in Euclid's ElementsIn Euclid's Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines. 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 Theorem)

The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.

The area of any square is equal to the product of two of its sides.

The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).

The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.

The proof is as follows:

Illustration including the new linesLet 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.

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 and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.

Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.

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 appears in Euclid's Elements as that of Proposition 1.47.[8]

 Garfield's proof

James A. Garfield (later President of the United States) is credited with a novel algebraic proof[9] using a trapezoid containing two examples of the triangle, the figure comprising one-half of the figure using four triangles enclosing a square shown below.

Proof using area subtraction.

 Similarity proof

From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.

 Proof by rearrangement

Proof of Pythagorean theorem by rearrangement of 4 identical right triangles. Since the total area and the areas of the triangles are all constant, the total black area is constant. But this can be divided into squares delineated by the triangle sides a, b, c, demonstrating that a2 + b2 = c2 .A proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (a + b)2. In both, the area of four identical triangles is removed. The remaining areas, a2 + b2 and c2, are equal. Q.E.D.

Animation showing another proof by rearrangement.

Proof using rearrangement.

A square created by aligning four right angle triangles and a large square.This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself (see Lebesgue measure and Banach-Tarski paradox). Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).

A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.[10]

 Algebraic proof

An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C.

The A-side angle and B-side angle of each of these triangles are complementary angles, 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:

However, as the large square has sides of length A + B, we can also calculate its area as (A + B)2, which expands to A2 + 2AB + B2.

(Distribution of the 4)

(Subtraction of 2AB)

 Proof by differential equations

One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.[11]

Proof using differential equations.As a result of a change da in side a,

by similarity of triangles and for differential changes. So

upon separation of variables.

which results from adding a second term for changes in side b.

Integrating gives

When a = 0 then c = b, so the "constant" is b2. So

As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral.

These quantities da and dc are respectively infinitely small changes in a and c. But we use instead real numbers Δa and Δc, then the limit of their ratio as their sizes approach zero is da/dc, the derivative, and also approaches c/a, the ratio of lengths of sides of triangles, and the differential equation results.

 Converse

The converse of the theorem is also true:

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.

This converse also appears in Euclid's Elements. It can be proven using the law of cosines (see below under Generalizations), or by the following proof:

Let ABC be a triangle with side lengths a, b, and c, with a2 + b2 = c2. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the