TUGAS I A

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. The theorem can also be generalized from a plane triangle to a trirectangular tetrahedron, in which case it is known as de Gua's theorem. The various proofs of the Pythagorean theorem all seem to require application of some version or consequence of the parallel postulate: proofs by dissection rely on the complementarity of the acute angles of the right triangle, proofs by shearing rely on explicit constructions of parallelograms, proofs by similarity require the existence of non-congruent similar triangles, and so on (S. Brodie). Based on this observation, S. Brodie has shown that the parallel postulate is equivalent to the Pythagorean theorem.
After receiving his brains from the wizard in the 1939 film The Wizard of Oz, the Scarecrow recites the following mangled (and incorrect) form of the Pythagorean theorem, "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." In the fifth season of the television program The Simpsons, Homer J. Simpson repeats the Scarecrow's line (Pickover 2002, p. 341). In the Season 2 episode "Obsession" (2006) of the television crime drama NUMB3RS, Charlie's equations while discussing a basketball hoop include the formula for the Pythagorean theorem.

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

Another proof by dissection is due to Perigal (left figure; Pergial 1873; Dudeney 1958; Madachy 1979; Steinhaus 1999, pp. 4-5; Ball and Coxeter 1987). A related proof is accomplished using the above figure at right, in which 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
4(1/2 ab) + d^2 (2)
2ab + (b-a)^2 (3)
2ab + b^2 - 2ab + a^2 (4)
a^2 + b^2 (5)
c^2 (6)
The Indian mathematician Bhaskara constructed a proof using the above figure, and another beautiful dissection proof is shown below (Gardner 1984, p. 154).
c^2 + 4(1/2 ab) = (a + b)^2 (7)
c^2 + 2ab = a^2 + 2ab + b^2 (8)
c^2 = a^2 + b^2 (9)

Several beautiful and intuitive proofs by shearing exist (Gardner 1984, pp. 155-156; Project Mathematics!).
Perhaps the most famous proof of all times is Euclid's geometric proof (Tropfke 1921ab; Tietze 1965, p. 19), although it is neither the simplest nor the most obvious. Euclid's proof used the figure below, which is sometimes known variously as the bride's chair, peacock tail, or windmill. The philosopher Schopenhauer has described this proof as a "brilliant piece of perversity" (Schopenhauer 1977; Gardner 1984, p. 153).

Let triangle ABC be a right triangle,squere CAFG,squere CBKH, squere ABED and CL parallel BE be squares,triangle FAB and triangle CAD. The triangles and are equivalent except for rotation, so
2 triangle FAB = 2 triangle CAD (10)
Shearing these triangles gives two more equivalent triangles
2 triangle CAD = ADLM (11)
Therefore,
squere ACGF = ADLM (12)
Similarly,
Squere BC = 2 triagle ABK (13)
so
a^2 + b^2 = cx + c(c-x) = c^2 (14)
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 = 1/4 root of 2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (15)
and equating to the area
K = 1/2 AB (16)
gives
1/4 a^2b^2 = 1/16 [2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (17)
Rearranging and simplifying gives
a^2 + b^2 = c^2 (18)
the Pythagorean theorem, where is the area of a triangle with sides a, b,c and (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 (Gardner 1984, pp. 155 and 161; Pappas 1989, pp. 200-201; Bogomolny). 1/2 [bases * altitude] (19)
1/2 (a + b) (a + b) (20)
1/2 ab + 1/2 ab + 1/2 c^2 (21)
Rearranging,
1/2 (a^2 + 2ab + b^2) = ab + 1/2 c^2 (22)
a^2 + 2ab + b^2 = 2ab + C^2 (23)
a^2 + b^2 = c^2 (24)
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^2 + bi = ce^-18 (25)
Taking the complex conjugate gives
a^2 - bi = ce^-18 (26)
Multiplying (25) by (26) gives
a^2 + b^2 = c^2 (27)
(Machover 1996).
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 ,d and (small triangle in the left figure; reproduced in the right figure) is similar to the right triangle with sides d , b ,y and (large triangle in the left figure; reproduced in the middle figure), giving
x/a = a/c y/b = b/c (28)
x = a^2/c y = b^2/c (29)
so
c = x + y = a^2/c + b^2/c = (a^2 + b^2)/c (30)
c^2 = a^2 + b^2 (31)
(Gardner 1984, p. 155 and 157). 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.
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