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© Copyright 1997, Jim Loy
The Pythagorean Theorem states:
In a right triangle, with sides (legs) a and b, and hypotenuse c, then c²=a²+b².
A right triangle is a triangle with one right angle (an angle of 90°). Its hypotenuse is the side opposite the right angle.
Proof #1: The simplest proof is an algebraic proof using similar triangles ABC, CBX, and ACX (in the diagram):
Since corresponding parts of similar triangles are proportional, a/x=c/a or a²=cx. And b/(c-x)=c/b or b²=c²-cx or c²=cx+b². Substituting a² for cx, we get c²=a²+b². Which is what we were trying to prove.
This proof is by Legendre, and was probably originally devised by an ancient Hindu mathematician. Euclid's proof is quite a bit more complicated than that. It is actually surprising that he did not come up with a proof similar to the above. But, his proof is clever, as well.
Proof #2: Here is another nice proof:
We start with a right triangle (in gold, in the diagram) with sides a, b, and c. We then build a big square, out of four copies of our triangle, as shown at the left. We end up with a square, in the middle, with sides c (we can easily show that this is a square).
We now construct a second big square, with identical triangles which are arranged as in the lower part of the diagram. This square has the same area as the square above it.
We now sum up the parts of the two big squares:
Area=2ab + c²
Area=2ab + a² + b²
These two areas are equal:
2ab + c²=2ab + a² + b²
c²=a² + b²
Proof #3: This diagram might look familiar. I've just drawn the squares on the sides of our right triangle. And, I've drawn a line from the right angle of the triangle, perpendicular to the hypotenuse, through the square which is on the hypotenuse. The idea is to prove that the little square (in blue) has the same area as the little rectangle (also in blue). I've named the width of this rectangle, x.
Using similar triangles, within our right triangle:
The area of the blue rectangle is xc. So we just plug in a²/c for x:
So, the area of the blue rectangle is equal to the area of the blue square. Similarly, the area of the other rectangle (to the left of the blue one) is equal in area of the other square (b²).
So, the total area of the big square is:
c²=a² + b²
That was actually fairly easy. I made that proof up, as I studied Euclid's proof, below.
Proof #4: We now look at Euclid's proof. Although he proved many of the theorems of algebra in his books, he did not have the methods of algebra (the manipulation of symbols) available to him. So, we must add a few more lines. We label a few points (see the diagram), and draw lines BI and CE.
Triangle ACE is congruent to triangle AIB, by the side-angle-side theorem (it's a postulate, nowadays), because AC=AI, AE=AB and angle CAE=angle IAB (both angles are congruent to angle CAB + a right angle). [Note: I am using the=sign for congruence. And I am labelling a line segment as AC, because I lack more appropriate symbols].
But, the area of triangle AIB is equal to half the area of square ACHI (They have the same base AI and height IH), and the area of triangle ACE is equal to half the area of the rectangle AGFE. So, the square ACHI is equal, in area, to the rectangle AGFE. Likewise, we show that the area of the smaller square is equal to the area of the smaller rectangle.
So, the area of the large square is equal to the sums of the areas of the two smaller squares. So, we've proved the Pythagorean theorem, again.
Here is a square with two diagonals. The area of the four triangles is 2a². The area of the square is c². Obviously a²+a²=c², a special case of the Pythagorean Theorem. If a is 1, then c is sqrt(2) [the square root of 2]. Or if c is 1, then a is sqrt(2)/2. This diagram does not seem to help in proving the Pythagorean Theorem.
Mathematics historians are fairly certain that the Pythagorean Theorem was known before Pythagoras. Various cultures seem to have known the length of the diagonal of a square (sqrt(2)) fairly accurately, more accurately than they could measure with a ruler. See The 3-4-5 Right Triangle In Ancient Egypt, where I express the opinion that knowledge of a 3-4-5 right triangle does not imply knowledge of the Pythagorean Theorem.
The book The Pythagorean Proposition, By Elisha Scott Loomis, is a fairly amazing book. It contains 256 proofs of the Pythagorean Theorem. It shows that you can devise an infinite number of algebraic proofs, like the first proof above. It shows that you can devise an infinite number of geometric proofs, like Euclid's proof. And it shows that there can be no proof using trigonometry, analytic geometry, or calculus. The book is out of print, by the way.
The Pythagorean Theorem depends on the parallel postulate. So, it does not hold for the non-Euclidean geometries. In relation to the above proofs
We also have the problem that there are no squares in the non-Euclidean geometries.
Note: Congruence applies to geometric figures, including line segments and angles. Equality applies to numbers, such as areas, lengths of line segments, and the measures of angles. A square and a rectangle are not equal, their areas are equal.
See Rik Littlefield's proof, which is interesting. Also, see The 3-4-5 Right Triangle In Ancient Egypt. The Pythagorean Theorem is also mentioned in Isosceles Triangles.
Proof #5: This proof is attributed to President Garfield. Given the right triangle at the bottom of this diagram, duplicate it as shown, and complete the trapezoid as shown. The larger triangle (with two sides c) is a right triangle (fairly easy to show). The areas of the three triangles are ab/2, ab/2 and c²/2. The sum of these is ab+c²/2. The area of a trapezoid is 1/2 the sum of the parallel sides times the height. Here that is (a+b)(a+b)/2 or (a²+2ab+b²)/2. These two expressions (for the three triangles and the trapezoid) represent the same area:
2ab+c²=a²+2ab+b² [double both sides]
This proof has a great deal in common with our second proof, above.
Proof #6: This proof is attributed to Bhaskara, a Hindu mathematician of the 12th century. We are given the bottom right triangle. We construct a square by making three copies of the triangle, as shown. We assume that a is the smallest side, less than or equal to b. The area of the large square is c². The side of the small square is b-a, and its area is (b-a)² or b²-2ab+a². The area of our triangle is ab/2. The area of all four triangles is 2ab. Then the area of all four triangles, plus the area of the small square is b²+a². So c²=a²+b².
This proof also has a great deal in common with our second proof, above.
Pythagorean triples: Pythagorean triples are sets of three integers (a, b, and c) which satisfy the equation c²=a²+b². In other words, they can be the sides of a right triangle. The smallest non-trivial triple is 3, 4, 5. Another is 5, 12, 13. Triples can be multiplied by any integer to produce another triple. For example, 3, 4, 5 can be doubled to produce 6, 8, 10, which is a Pythagorean triple (and a triangle that is similar to the 3-4-5 right triangle). It is possible to devise a formula for producing Pythagorean triples. [Note: Trivial triples are like 0, 0, 0 or 0, 1, 1.] See Pythagorean Triples.
The study of Pythagorean triples leads to Fermat's Last Theorem, that cx=ax+bxhas no non-trivial solutions with integers for x>2.
In trigonometry ("trig" for short), which I will eventually write about at greater length, deals with angles and certain lengths and their ratios. In the right triangle at the left, the sine of angle A (abbreviated "sin a") is defined as a/c. All right triangles with the same angle A are similar (as shown by Euclid), so this ratio (a/c) is the same regardless of the size of the triangle. Similarly, the cosine of angle A is b/c. One fact (as well as a few other facts) of trig follows directly from the Pythagorean Theorem: for any angle x, sin² x + cos² x=1. To prove it, just plug in a/c for sin x and b/c for cos x, and the sum comes out to 1.
In a general triangle, not necessarily a right triangle, there is an equation similar to the Pythagorean Theorem: c²=a² + b² - 2ab cos C. This is called the Law of Cosines. Similarly, a²=b² + c² - 2bc cos A and b²=a² + c² - 2ac cos B. These can be proved in more than one way, using the Pythagorean theorem.
See My Three Triangle Puzzle.
Here are a few simple problems which use the Pythagorean Theorem:
Here are some answers and comments:
Proof #7:Here is another proof, similar to proof #2, and some of the others. The area of the figure on the left is a² + b² (it is made up of two squares). By rearranging the three pieces, we get the square on the right, with area c². This also shows us how to dissect two squares of any size and put the pieces back together to form one square (or vice versa).
The drawing on the right is similar to one found in the amazing book, 1000 PlayThinks by Ivan Moscovich. The book uses it as a dissection problem (using pieces of the smaller squares to make up the largest square). Then it says, "If you can do [this], what have you done?" The answer in the back of the book says, "Congratulations, you have demonstrated the truth behind the Pythagorean theorem." I have two problems with that. One is that (in mathematics) "demonstrate" means "prove," which we have not done. And second, the triangle in the diagram is not a right triangle (look closely at the "right angle," which is about 92 degrees here). If anything, it demonstrates the falsity of the Pythagorean theorem. Of course, it does neither, as the diagram was drawn sloppily. I intentionally drew mine sloppily. So be careful about "proving" things by drawing them.
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