Pythagorean Theorem

For a Right Triangle with legs $a$ and $b$ and Hypotenuse $c$,

$$a^2+b^2=c^2.$$ (1)

Many different proofs exist for this most fundamental of all geometric theorems.

A clever proof by Dissection which reassembles two small squares into one larger one was given by the Arabian mathematician Thabit Ibn Qurra (Ogilvy 1994, Frederickson 1997).

Another proof by Dissection is due to Perigal (Pergial 1873, Dudeney 1970, Madachy 1979, Ball and Coxeter 1987).

The Indian mathematician Bhaskara constructed a proof using the following figure.

Another beautiful dissection proof is shown below.

$$c^2+4({\textstyle{1\over 2}}ab)=(a+b)^2$$ (2)

$$c^2+2ab=a^2+2ab+b^2$$ (3)

$$c^2=a^2+b^2.$$ (4)

In the above figure, the Area of the large Square is four times the Area of one of the Triangles plus the Area of the interior Square. From the figure, $d=b-a$, so

$$A = 4({\textstyle{1\over 2}}ab)+d^2= 2ab+(b-a)^2 = 2ab+b^2-2ab+a^2$$

$$= a^2+b^2=c^2.$$ (5)

Perhaps the most famous proof of all times is Euclid's geometric proof. Euclid's proof used the figure below, which is sometimes known variously as the Bride's Chair, Peacock's Tail, or Windmill.

Let $\Delta ABC$ be a Right Triangle, $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern6.4pt \vrule width.6pt} \hrule height.6pt}CAFG$, $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern6.4pt \vrule width.6pt} \hrule height.6pt}CBKH$, and $\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern6.4pt \vrule width.6pt} \hrule height.6pt}ABED$ be squares, and $CL\vert\vert BD$. The Triangles $\Delta FAB$ and $\Delta CAD$ are equivalent except for rotation, so

$$2\Delta FAB=2\Delta CAD.$$ (6)

Shearing these Triangles gives two more equivalent Triangles

$$2\Delta CAD=\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}ADLM.$$ (7)

Therefore,

$$\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \ker... ...eight6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}ADLM.$$ (8)

Similarly,

$$\vbox{\hrule height.6pt\hbox{\vrule width.6pt height6pt \ker... ...t height6pt \kern10.6pt \vrule width.6pt} \hrule height.6pt}BL$$ (9)

so

$$a^2+b^2=cx+c(c-x)=c^2.$$ (10)

Heron proved that $AK$, $CL$, and $BF$ intersect in a point (Dunham 1990, pp. 48-53).

Heron's Formula for the Area of the Triangle, contains the Pythagorean theorem implicitly. Using the form

$$K={\textstyle{1\over 4}}\sqrt{2a^2b^2+2a^2c^2+ab^2c^2-(a^4+b^4+c^4)}$$ (11)

and equating to the Area

$$K={\textstyle{1\over 2}}ab$$ (12)

gives

$${\textstyle{1\over 4}}a^2b^2=2a^2b^2+2a^2c^2+ab^2c^2-(a^4+b^4+c^4).$$ (13)

Rearranging and simplifying gives

$$a^2+b^2=c^2,$$ (14)

the Pythagorean theorem, where $K$ is the Area of a Triangle with sides $a$, $b$, and $c$ (Dunham 1990, pp. 128-129).

A novel proof using a Trapezoid was discovered by James Garfield (1876), later president of the United States, while serving in the House of Representatives (Pappas 1989, pp. 200-201; Bogomolny).

$$A_{\rm trapezoid} = {\textstyle{1\over 2}}\sum {\rm [bases]}\ \cdot {\rm [altitude]}$$

$$= {\textstyle{1\over 2}}(a+b)(a+b)$$

$$= {\textstyle{1\over 2}}ab+{\textstyle{1\over 2}}ab+{\textstyle{1\over 2}}c^2.$$ (15)

Rearranging,

$${\textstyle{1\over 2}}(a^2+2ab+b^2)=ab+{\textstyle{1\over 2}}c^2$$ (16)

$$a^2+2ab+b^2=2ab+c^2$$ (17)

$$a^2+b^2=c^2.$$ (18)

An algebraic proof (which would not have been accepted by the Greeks) uses the Euler Formula. Let the sides of a Triangle be $a$, $b$, and $c$, and the Perpendicular legs of Right Triangle be aligned along the real and imaginary axes. Then

$$a+bi=ce^{i\theta}.$$ (19)

Taking the Complex Conjugate gives

$$a-bi=ce^{-i\theta}.$$ (20)

Multiplying (19) by (20) gives

$$a^2+b^2=c^2.$$ (21)

Another algebraic proof proceeds by similarity.

It is a property of Right Triangles, such as the one shown in the above left figure, that the Right Triangle with sides $x$, $a$, and $d$ (small triangle in the left figure; reproduced in the right figure) is similar to the Right Triangle with sides $d$, $b$, and $y$ (large triangle in the left figure; reproduced in the middle figure), giving

$${x\over a}={a\over c} \qquad {y\over b}={b\over c}$$ (22)

$$x={a^2\over c}\qquad y={b^2\over c},$$ (23)

so

$$c\equiv x+y={a^2\over c}+{b^2\over c} = {a^2+b^2\over c}$$ (24)

$$c^2=a^2+b^2.$$ (25)

Because this proof depends on proportions of potentially Irrational Numbers and cannot be translated directly into a Geometric Construction, it was not considered valid by Euclid.

See also Bride's Chair, Cosines Law, Peacock's Tail, Pythagoras's Theorem, Windmill

References

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Dixon, R. ``The Theorem of Pythagoras.'' §4.1 in Mathographics. New York: Dover, pp. 92-95, 1991.

Dudeney, H. E. Amusements in Mathematics. New York: Dover, p. 32, 1958.

Dunham, W. ``Euclid's Proof of the Pythagorean Theorem.'' Ch. 2 in Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, 1990.

Frederickson, G. Dissections: Plane and Fancy. New York: Cambridge University Press, pp. 28-29, 1997.

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Ogilvy, C. S. Excursions in Mathematics. New York: Dover, p. 52, 1994.

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Perigal, H. ``On Geometric Dissections and Transformations.'' Messenger Math. 2, 103-106, 1873.

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Shanks, D. Solved and Unsolved Problems in Number Theory, 4th ed. New York: Chelsea, pp. 123-127, 1993.

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Yancey, B. F. and Calderhead, J. A. ``New and Old Proofs of the Pythagorean Theorem.'' Amer. Math. Monthly 5, 73-74, 1898.

Yancey, B. F. and Calderhead, J. A. ``New and Old Proofs of the Pythagorean Theorem.'' Amer. Math. Monthly 6, 33-34 and 69-71, 1899.