the Pythagorean Theorem

Before embarking on trigonometry, there are a couple of things you need to know well about geometry, namely the Pythagorean theorem and similar triangles. Both of these are used over and over in trigonometry.

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The Pythagorean theorem states that the square of the hypotenuse is the sum of the squares of the other two sides, that is,

This theorem is useful to determine one of the three sides of a right triangle if you know the other two. For instance, if two legs are *a* = 5, and *b* = 12, then you can determine the hypotenuse *c* by squaring the lengths of the two legs (25 and 144), adding the two squares together (169), then taking the square root to get the value of *c,* namely, 13.

Likewise, if you know the hypotenuse and one leg, then you can determine the other. For instance, if the hypotenuse is *c* = 41, and one leg is *a* = 9, then you can determine the other leg *b* as follows. Square the hypotenuse and the first leg (1681 and 81), subtract the square of the first leg from the square of the hypotenuse (1600), then take the square root to get the value of the other leg *b,* namely 40.

*Proof*: Start with the right triangle *ABC* with right angle at *C.* Draw a square on the hypotenuse *AB,* and translate the original triangle *ABC* along this square to get a congruent triangle *A'B'C'* so that its hypotenuse *A'B'* is the other side of the square (but the triangle *A'B'C'* lies inside the square). Draw perpendiculars *A'E* and *B'F* from the points *A'* and *B'* down to the line *BC.* Draw a line *AG* to complete the square *ACEG.*

Note that *ACEG* is a square on the leg *AC* of the original triangle. Also, the square *EFB'C'* has side *B'C'* which is equal to *BC,* so it equals a square on the leg *BC.* Thus, what we need to show is that the square *ABB'A'* is equal to the sum of the squares *ACEG* and *EFB'C'.*

But that's pretty easy by cutting and pasting. Start with the big square *ABB'A'.* Translate the triangle *A'B'C'* back across the square to triangle *ABC,* and translate the triangle *AA'G* across the square to the congruent triangle *BB'F.* Paste the pieces back together, and you see you've filled up the squares *ACEG* and *EFB'C'.* Therefore, *ABB'A'* = *ACEG* + *EFB'C',* as required.

Q.E.D.

In fact, as Euclid showed, each of these two conditions implies the others. That is to say, if corresponding angles are equal, then the three ratios are equal (Prop. VI.4), but if the three ratios are equal, then corresponding angles are equal (Prop. VI.5). Thus, it is enough to know either that their corresponding angles are equal or that their sides are proportional in order to conclude that they are similar triangles.

Typically, the smaller of the two similar triangles is part of the larger. For example, in the diagram to the left, triangle *AEF* is part of the triangle *ABC,* and they share the angle *A.* When this happens, the opposite sides, namely *BC* and *EF,* are parallel lines.

This situation frequently occurs in trigonometry applications, and for many of those, one of the three angles *A, B,* or *C* is a right angle.