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 + c²These two areas are equal:
Area=2ab + a² + b²
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:
a/c=x/aThe area of the blue rectangle is xc. So we just plug in a²/c for x:
x=a²/c
area=(a²/c)c=a²So, 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 + b²That 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
- There are no similar triangles in the non-Euclidean geometries.
- We cannot prove that the above triangles have half the area of the square or rectangle.
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.
Addendum #1:
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:ab+c²/2=(a²+2ab+b²)/2This proof has a great deal in common with our second proof, above.
2ab+c²=a²+2ab+b² [double both sides]
c²=a²+b²
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.
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