Proof #1: Looking at the areas of triangles and squares
Consider the following right triangle:
c
a
b
Using that triangle, we can create the “square of c,” and get the following picture:
Since each side of the square “c” can be thought of as the hypotenuse of the original right
triangle, we can draw 3 more corresponding right triangles and get the following picture:
We can view this picture in a number of ways. The first way is to view it as a square. If we do
so, then to compute the area of the square we would just multiply the length of the two sides
together. The length of a side of the above square is a  b. Therefore the area of the above
square is a  b a  b  a 2  2ab  b 2 .
Another way to view the above picture is 4 triangles and one square, each of which we can
compute the area. The area of the square is c 2 , and the area of each triangle is
1
2
1
ab . Summing
2


up the areas of the square and 4 triangles and the one square we get 4 ab   c 2  2ab  c 2 .
Now, even though we viewed the above picture in two different ways and came up with two
different expressions of the area, the two areas are equivalent. Therefore, we can set the two
expressions equal. a 2  2ab  b 2  2ab  c 2  a 2  b 2  c 2 . QED
Proof #2: Using similarity of triangles
Consider the following picture, where “d” is an altitude and x  y  c.
Looking at the above picture, we can see that we have three separate triangles: abc, dyb, and
adx.
We can show that each of the smaller triangles dyb and adx are similar to the larger triangle
abc by AAA. Looking at abc and dyb , we can see that those two are similar. They both
have right angles and they both share the same exact angle (where y and b intersect). Since they
have two corresponding congruent angles, the third corresponding angles must be congruent.
Therefore the two triangles are similar by AAA.
You can show that abc and adx are similar triangles by using an analogous argument.
Since the triangles are similar we can set up the following proportions:
x a
a2
 x
a
c
c
and
b c
b2
. Since x + y = c, then
 y
y b
c
b2 a2
x y c

 c  b 2  a 2  c 2 . QED
c
c