The Pythagorean theorem does in fact have MANY mathematical applications and uses. Due to these far-reaching effects, some mathematicians consider the theorem to be one of THE most important in all of mathematics. We hope the application of the theorem below serves to reinforce this idea of the Pythagorean theorem as an elegant, useful one. Without further ado, we now approach the task at hand...
First, we start with two SIMILAR triangles, abc and ABC, with ABC being the larger of the two (see Diagram 1 below) Our goal is to show that B/b = A/a = C/c, using both the given information and the Pythagorean theorem. As indicated in the diagram, since the two triangles are similar, we can place the smaller triangle over the top of the larger one, resulting in the image below, with the various labelled dimensions.
Diagram 1
The proof now proceeds in a two parts:
Part 1. Looking at diagram 1, we see that
(1/2) AB = (1/2) ab + b
(A - a) + (1/2) (A-a) (B - b)
= (1/2) ab + bA - ab + (1/2) AB - (1/2) Ab - (1/2) aB + (1/2) ab, so that
0 = (1/2) ab - ab + (1/2)ab + Ab - (1/2) Ab - (1/2) aB.
Cancelling terms, we have
0 = (1/2) Ab - (1/2) aB, which implies that
aB = Ab. Dividing both sides by ab and cancelling, we find
B/b = A/a.
We shall designate these ratios r, so that r = B/b = A/a. In other words, we have that A = ar, and that B = br, useful notations for Part 2 below.
Part 2. (The Pythagorean theroem makes its appearance!!) The Pythagorean theorem states for triangles abc and ABC the following:
a2 + b2
= c2, (equation
1), and
A2 + B2
= C2. (equation 2)
Substituting in (ar) for A and (br) for B in equation 2 above gives
This completes the proof, as we have shown that r = A/a = B/b = C/c.