6.2 The Pythagorean Theorem
Without any question, the most famous "named" theorem in mathematics is the Pythagorean Theorem, the subject of this section. There are literally hundreds of known proofs of this theorem; they take many forms and have a variety of emphases. We often think of the theorem in algebraic terms — under particular circumstances, $a^2+b^2=c^2$. Indeed, our proofs in this section will have an algebraic flavor. However, it would seem that to the ancient Greeks, the theorem was about physical squares and their areas. As we will see in the exercises in Chapter 8, Euclid's proof of the Pythagorean Theorem demonstrated how to break the square drawn on the hypotenuse of a right triangle into two rectangles, one of which had the same area as the square drawn on one leg of the triangle and the other of which had the same area as the square drawn on the other leg.
Pythagoras settled in a small Greek city in southern Italy called Crotona. Here he organized a secret society or brotherhood whose members were called Pythagoreans. The discussions and ideas of the Pythagoreans are responsible for many important ideas in mathematics and astronomy.
- Theorem 6.2:
The Pythagorean Theorem
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
Given: $\triangle
ABC$ is a right triangle with right angle $\angle ACB$. Prove: $c^2=a^2+b^2$ |
We will consider two very different proofs of this theorem. The first makes use of the ideas of the last
section, in which the altitude to the hypotenuse of a right triangle led to geometric mean results. The second is an area proof, but still is fairly algebraic. In Chapter 8, we will see the approach Euclid took in proving this pivotal theorem.
Proof 1: Draw an altitude to the hypotenuse of the right triangle. Then $b$ is the geometric mean between and $c$ and $d$. Also, $a$ is the geometric mean between $c$ and $e$. Therefore, $a^2=c\cdot e$ and $b^2=c\cdot d$.
Proof 2: Make four exact copies of $\triangle ABC$ and arrange them as shown to the right. Since the acute angles of a right triangle are complementary, the figure formed has four right angles and sides of length $c$. This is a square, and its area is $c^2$. But this area is also made up of the smaller center square and four right triangles. The center square has an area of $(a-b)^2$. The four triangles have a total area of $4\left(\frac{1}{2}a\cdot b\right)$.
\begin{align}
c^2&=(a-b)^2+4\left(\frac{1}{2}a\cdot b\right)\\
c^2&=\left(a^2-2ab+b^2\right)+\left(2a\cdot b\right)\\
c^2&=a^2+b^2
\end{align}
It is critical to remember that, like all theorems, there is a given situation for which the Pythagorean Theorem holds. The theorem does not say, "$a^2+b^2=c^2$." It says that "In a right triangle with legs $a$ and $b$ and hypotenuse $c$, $a^2+b^2=c^2$ ." There is a hypothesis (specifically, the triangle is a right triangle) that is crucial.
As you do problems with right triangles, you should keep one obvious fact in mind. Since the largest angle of a right triangle is the right angle — since in a right triangle with hypotenuse $c$, $c^2$ is the sum of two positive numbers — the longest side of a right triangle is necessarily the hypotenuse. It is ridiculous to give an answer to a problem and have the hypotenuse shorter than one of the legs.
The Pythagorean theorem can be generalized to include figures other than squares. In fact, Hippocrates of Chios in the fifth century BC, used similar figures constructed on each of the three sides.
If one erects similar figures with
corresponding sides on the sides of a right triangle, then the sum of
the areas of the ones on the two samller sides equals the area of the
one on the larger side. (Heath, T. L., A History of Greek Mathematics,
Oxford University Press, 1921; reprinted by Dover, 1981.)
Marie El-Nabbout, 11 April 2014, Created with GeoGebra