# Pythagorean theorem

## Relation between sides of a right triangle / From Wikipedia, the free encyclopedia

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In mathematics, the **Pythagorean theorem** or **Pythagoras' theorem** is a fundamental relation in Euclidean geometry between the three sides of a right triangle. It states that 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 on the other two sides.

**Quick Facts**Type, Field ...

Type | Theorem |
---|---|

Field | Euclidean geometry |

Statement | 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). |

Symbolic statement | $a^{2}+b^{2}=c^{2}$ |

Generalizations | |

Consequences |

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

- $a^{2}+b^{2}=c^{2}.$

The theorem is named for the Greek philosopher Pythagoras, born around 570 BC. The theorem has been proved numerous times by many different methods – possibly the most for any mathematical theorem. The proofs are diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years.

When Euclidean space is represented by a Cartesian coordinate system in analytic geometry, Euclidean distance satisfies the Pythagorean relation: the squared distance between two points equals the sum of squares of the difference in each coordinate between the points.

The theorem can be generalized in various ways: to higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and 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.^{[citation needed]}

### Rearrangement proofs

In one rearrangement proof, two squares are used whose sides have a measure of $a+b$ and which contain four right triangles whose sides are * a*,

*and*

**b***, with the hypotenuse being*

**c***. In the square on the right side, the triangles are placed such that the corners of the square correspond to the corners of the right angle in the triangles, forming a square in the center whose sides are length*

**c***. Each outer square has an area of $(a+b)^{2}$ as well as $2ab+c^{2}$, with $2ab$ representing the total area of the four triangles. Within the big square on the left side, the four triangles are moved to form two similar rectangles with sides of length*

**c***and*

**a***. These rectangles in their new position have now delineated two new squares, one having side length*

**b***is formed in the bottom-left corner, and another square of side length*

**a***formed in the top-right corner. In this new position, this left side now has a square of area $(a+b)^{2}$ as well as $2ab+a^{2}+b^{2}$. Since both squares have the area of $(a+b)^{2}$ it follows that the other measure of the square area also equal each other such that $2ab+c^{2}$ = $2ab+a^{2}+b^{2}$. With the area of the four triangles removed from both side of the equation what remains is $a^{2}+b^{2}=c^{2}.$*

**b**^{[2]}

In another proof rectangles in the second box can also be placed such that both have one corner that correspond to consecutive corners of the square. In this way they also form two boxes, this time in consecutive corners, with areas $a^{2}$ and $b^{2}$which will again lead to a second square of with the area $2ab+a^{2}+b^{2}$.

English mathematician Sir Thomas Heath gives this proof in his commentary on Proposition I.47 in Euclid's *Elements*, and mentions the proposals of German mathematicians Carl Anton Bretschneider and Hermann Hankel that Pythagoras may have known this proof. Heath himself favors a different proposal for a Pythagorean proof, but acknowledges from the outset of his discussion "that the Greek literature which we possess belonging to the first five centuries after Pythagoras contains no statement specifying this or any other particular great geometric discovery to him."^{[3]} Recent scholarship has cast increasing doubt on any sort of role for Pythagoras as a creator of mathematics, although debate about this continues.^{[4]}

### Algebraic proofs

The theorem can be proved algebraically using four copies of the same triangle arranged symmetrically around a square with side *c*, as shown in the lower part of the diagram.^{[5]} 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

- $c^{2}=(b+a)^{2}-2ab=b^{2}+2ab+a^{2}-2ab=a^{2}+b^{2}.$

A similar proof uses 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.^{[6]} The triangles are similar with area ${\tfrac {1}{2}}ab$, while the small square has side *b* − *a* and area (*b* − *a*)^{2}. The area of the large square is therefore

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

But this is a square with side *c* and area *c*^{2}, so

- $c^{2}=a^{2}+b^{2}.$

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.^{[7]}

### Proof using similar triangles

^{2}denotes the square of the length AB and not the product $A\times B^{2}.$

This proof is based on the proportionality of the sides of three 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 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:

- ${\frac {BC}{AB}}={\frac {BH}{BC}}{\text{ and }}{\frac {AC}{AB}}={\frac {AH}{AC}}.$

The first result equates the cosines of the angles *θ*, whereas the second result equates their sines.

These ratios can be written as

- $BC^{2}=AB\times BH{\text{ and }}AC^{2}=AB\times AH.$

Summing these two equalities results in

- $BC^{2}+AC^{2}=AB\times BH+AB\times AH=AB(AH+BH)=AB^{2},$

which, after simplification, demonstrates the Pythagorean theorem:

- $BC^{2}+AC^{2}=AB^{2}.$

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.^{[8]}

### Einstein's proof by dissection without rearrangement

Albert Einstein gave a proof by dissection in which the pieces do not need to be moved.^{[9]} Instead of using a square on the hypotenuse and two squares on the legs, one can use any other shape that includes the hypotenuse, and two similar shapes that each include one of two legs instead of the hypotenuse (see Similar figures on the three sides). In Einstein's proof, the shape that includes the hypotenuse is the right triangle itself. The dissection consists of dropping a perpendicular from the vertex of the right angle of the triangle to the hypotenuse, thus splitting the whole triangle into two parts. Those two parts have the same shape as the original right triangle, and have the legs of the original triangle as their hypotenuses, and the sum of their areas is that of the original triangle. Because the ratio of the area of a right triangle to the square of its hypotenuse is the same for similar triangles, the relationship between the areas of the three triangles holds for the squares of the sides of the large triangle as well.

### Trigonometric proof using Einstein's construction

Both the proof using similar triangles and Einstein's proof rely on constructing the height to the hypotenuse of the right triangle $\triangle ABC$. The same construction provides a trigonometric proof of the Pythagorean theorem using the definition of the sine as a ratio inside a right triangle:

- $\sin \alpha ={\frac {a}{c}},$
- $\sin \beta ={\frac {b}{c}},$
- $c=b\sin \beta +a\sin \alpha ={\frac {b^{2}}{c}}+{\frac {a^{2}}{c}},$

and thus

- $c^{2}=a^{2}+b^{2}.$

This proof is essentially the same as the above proof using similar triangles, where some ratios of lengths are replaced by sines.
^{[citation needed]}

### Euclid's proof

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.^{[10]}

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.
^{[11]} - 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.
- 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, BD is equal to BC and angle ABD equals angle FBC, 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, and this line is parallel to FB, then square BAGF must be twice in area to triangle FBC.
- Therefore, rectangle BDLK must have the same area as square BAGF = AB
^{2}. - By applying steps 3 to 10 to the other side of the figure, it can be similarly shown that rectangle CKLE must have the same area as square ACIH = AC
^{2}. - Adding these two results, AB
^{2}+ AC^{2}= BD × BK + KL × KC - Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
- Therefore, AB
^{2}+ AC^{2}= BC^{2}, since CBDE is a square.

This proof, which appears in Euclid's *Elements* as that of Proposition 47 in Book 1, demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.^{[12]}^{[13]}
This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.^{[14]}^{[15]}

### Proofs by dissection and rearrangement

Another by rearrangement is given by the middle animation. A large square is formed with area *c*^{2}, 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 *a*^{2} and *b*^{2}, which must have the same area as the initial large square.^{[16]}

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 way of cutting one figure into pieces and rearranging them to get another figure is called dissection. This shows the area of the large square equals that of the two smaller ones.^{[17]}

### Proof by area-preserving shearing

As shown in the accompanying animation, area-preserving shear mappings and translations can transform the squares on the sides adjacent to the right-angle onto the square on the hypotenuse, together covering it exactly.^{[18]} Each shear leaves the base and height unchanged, thus leaving the area unchanged too. The translations also leave the area unchanged, as they do not alter the shapes at all. Each square is first sheared into a parallelogram, and then into a rectangle which can be translated onto one section of the square on the hypotenuse.

### Other algebraic proofs

A related proof was published by future U.S. President James A. Garfield (then a U.S. Representative) (see diagram).^{[19]}^{[20]}^{[21]} 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.^{[22]}^{[23]}^{[24]}

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.

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:

- ${\frac {dy}{dx}}={\frac {x}{y}}.$

This can be rewritten as $y\,dy=x\,dx$ , which is a differential equation that can be solved by direct integration:

- $\int y\,dy=\int x\,dx\,,$

giving

- $y^{2}=x^{2}+C.$

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*.

The converse of the theorem is also true:^{[25]}

Given a triangle with sides of length

a,b, andc, ifa^{2}+b^{2}=c^{2}, then the angle between sidesaandbis a right angle.

For any three positive real numbers *a*, *b*, and *c* such that *a*^{2} + *b*^{2} = *c*^{2}, there exists a triangle with sides *a*, *b* and *c* as a consequence of the converse of the triangle inequality.

This converse appears in Euclid's *Elements* (Book I, Proposition 48): "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."^{[26]}

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

Let *ABC* be a triangle with side lengths *a*, *b*, and *c*, with *a*^{2} + *b*^{2} = *c*^{2}. 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* = √*a*^{2} + *b*^{2}, 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 proved without assuming the Pythagorean theorem.^{[27]}^{[28]}

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:^{[29]}

- If
*a*^{2}+*b*^{2}=*c*^{2}, then the triangle is right. - If
*a*^{2}+*b*^{2}>*c*^{2}, then the triangle is acute. - If
*a*^{2}+*b*^{2}<*c*^{2}, then the triangle is obtuse.

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

- sgn(
*α*+*β*−*γ*) = sgn(*a*^{2}+*b*^{2}−*c*^{2}),

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.^{[30]}