The Pythagorean Theorem

Abstract

The Pythagorean Theorem is one of the oldest, best known, and most useful theorems in all of mathematics, and it has also surely been proved in more different ways than any other. Euclid gave two proofs of it in the Elements, as Proposition I,47, and also as Proposition VI,31, a more general but less well-known formulation concerning arbitrary ‘figures’ described on the sides of a right triangle. The first of those demonstrations is based on a comparison of areas and the second on similarity theory, a basic distinction that can be used as a first step in classifying many other proofs of the theorem as well.

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The Pythagorean Theorem is one of the oldest, best known, and most useful theorems in all of mathematics, and it has also surely been proved in more different ways than any other. Euclid gave two proofs of it in the Elements, as Proposition I,47, and also as Proposition VI,31, a more general but less well-known formulation concerning arbitrary ‘figures’ described on the sides of a right triangle. The first of those demonstrations is based on a comparison of areas and the second on similarity theory, a basic distinction that can be used as a first step in classifying many other proofs of the theorem as well.

Since Euclid’s time hundreds of other proofs have been given — not because the correctness of the result or the rigor of Euclid’s arguments have ever been questioned, but principally because the theorem has fascinated generations of individuals, not only professional mathematicians but students and amateurs, who have felt challenged to apply their own ingenuity to prove it. The multitude of proofs thus created stands as an exemplar par excellence of the desire to find a previously undiscovered path to a goal.

Extensive compilations of proofs of the Pythagorean Theorem have appeared in several publications. Sources in English include a series of twelve articles by Benjamin F. Yanney and James A. Calderhead (Yanney and Calderhead 1896–9) that appeared in vols. 3 through 6 of the American Mathematical Monthly, each entitled “New and old proofs of the Pythagorean theorem”; the book (Loomis 1940) by Elisha S. Loomis, first published in 1927 and reprinted in 1968 by the National Council of Teachers of Mathematics; and the geometry web pages maintained by Alexander Bogomolny (Bogomolny 2012).

The first of those references nominally presents 100 different proofs; the second, 367; and the third, 96; but the qualifier ‘nominally’ is important for several reasons. First, as each of the compilers points out, some of the proofs admit numerous variations (sometimes thousands in the case of similarity arguments, depending on which particular sets of proportions are employed). Second, especially in Loomis’s book, distinctions among proofs are often not clearly or carefully made, despite his proclaimed intent to classify and arrange the proofs according to “method of proof and type of figure used.” Indeed, the very first of his ‘algebraic’ proofs — those based on similarity relations, as opposed to ‘geometric’ proofs based on area comparisons — appears to differ little, if at all, from his ‘algebraic’ proofs 38 and 93 (described as involving “the mean proportional principle” and “the theory of limits,” respectively), or from ‘geometric’ proof 230; and it is also similar to Euclid’s Proposition VI,31 — of which, incredibly, Loomis appears to have been completely unaware! Third, not all of the proofs given in Yanney’s and Calderhead’s articles or in Loomis’s book are correct! The most egregious example is Loomis’s ‘algebraic’ proof 16, apparently taken over uncritically from Yanney’s and Calderhead’s proof X. Ostensibly a proof by reductio, it assumes the Pythagorean Theorem to be true, derives a true consequence from it, and then declares the assumption to have been justified!¹ Several other fallacious proofs from those two earlier sources are also cited on Bogomolny’s web site.

Loomis’s book is problematical on other grounds as well. The very idea, for example, of distinguishing proofs according to the diagrams used to represent them seems fatally flawed, both because the same diagram, interpreted differently, may be used to represent conceptually distinct arguments, and because, conversely, some arguments can be represented by more than one distinct diagram. (See below for further discussion of both points.) Loomis’s criteria for excluding some proofs on the grounds that they are mere variants of ones given, while including others that seem hardly distinguishable from them, are vague, to say the least; and in several instances he also made categorical declarations with no, or with only weak, attempts at justification. Thus, for example, he declared the first of the ‘algebraic’ proofs he listed (the third of the proofs considered below) to be “the shortest proof possible,” without further discussion.² He also claimed (pp. viii and 224) that “no trigonometric proof [of the Pythagorean Theorem] is possible,” because “all the fundamental formulae of trigonometry are themselves based upon [the trigonometric form of] that theorem,” namely, the identity $\cos ^{2}\theta +\sin ^{2}\theta = 1$. But that is simply false: A very simple derivation of that identity, based directly on the ratio definitions for sine and cosine, is given below (proof 4).³

The purpose of the analyses that follow is not to duplicate the excellent commentaries on Bogomolny’s site (though all the proofs considered below are to be found there), nor to provide a critique of all, or even most, of the proofs collected by Yanney and Calderhead or Loomis. Rather, the aim of the case study undertaken in this chapter is the more modest one of examining and comparing seven proofs of the Pythagorean Theorem, selected as representative examples of distinct conceptual approaches that have been taken over the centuries. The two proofs given by Euclid⁴ provide a natural starting point.

• 1. Elements, Book I, Proposition 47: In right-angled triangles, the [area of the] square on the side subtending the right angle is equal to [the sum of the areas of] the squares on the sides containing the right angle.

FormalPara Proof summary:

The proof is based on the well-known ‘windmill’ or ‘bride’s chair’ diagram (Figure 5.1), constructed as follows:

Fig. 5.1

Euclid’s ‘windmill’ diagram

Given the right triangle ABC with right angle BAC, erect squares on each of its sides (as justified by the immediately preceding proposition I,46), and note that since angles BAC, BAG, and CAH are all right angles, $\overline{\mathit{AH}}$ and $\overline{\mathit{AG}}$ are extensions of the segments $\overline{\mathit{BA}}$ and $\overline{\mathit{CA}}$, so that segments $\overline{\mathit{CG}}$ and $\overline{\mathit{BH}}$ are parallel to segments $\overline{\mathit{BF}}$ and $\overline{\mathit{CK}}$, respectively. Next, drop a perpendicular from A to segment $\overline{\mathit{DE}}$, intersecting $\overline{\mathit{BC}}$ at M and $\overline{\mathit{DE}}$ at L. The segment $\overline{\mathit{LM}}$ then divides the square on $\overline{BC}$ into two rectangles, and the central idea of the proof is to show that the area of the square on $\overline{\mathit{AB}}$ is equal to the area of the rectangle BDLM and the area of the square on $\overline{\mathit{AC}}$ is equal to the area of the rectangle CELM. The principal tool for doing so is Proposition I,41, which states, in essence, that all triangles with the same base and altitude have the same area, or in other words, that the area of a triangle is invariant under the action of any shear transformation parallel to its base. Accordingly, if the segments $\overline{\mathit{AD}}$, $\overline{\mathit{AE}}$, $\overline{\mathit{CF}}$, and $\overline{\mathit{BK}}$ are drawn, then triangles BCF and ABF have the same area (half the area of the square on $\overline{\mathit{AB}}$), as do triangles BCK and ACK (half the area of the square on $\overline{\mathit{AC}}$); and likewise, triangle ABD has half the area of rectangle BDLM and triangle ACE half the area of rectangle CELM. But triangles BCF and ABD are congruent, as are triangles BCK and ACE, by the side-angle-side criterion (Proposition I,4), since angles CBF and DBA are equal (each being a right angle plus angle ABC), as are angles BCK and ACE (each being a right angle plus angle ACB). Hence half the area of the square on $\overline{\mathit{AB}}$ plus half the area of the square on $\overline{\mathit{AC}}$ is equal to half the area of rectangle BDLM plus half the area of rectangle CELM, from which the desired result follows by multiplying by two.

• 2. Elements, Book VI, Proposition 31: In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. (See Figure 5.2.)

  Fig. 5.2

  

  Euclid’s Proposition VI,31

FormalPara Overview of proof:

Although its statement is much more general than that of Proposition I,47, the proof of Proposition VI,31 is much simpler. Indeed, the proof follows by inspection from Figure 5.3 below, once the following facts are recognized:

Fig. 5.3

Three similar triangles

• (a) The statement of the proposition does not require that the figures described on the sides of the triangle must be exterior to the triangle, nor that they not overlap.

• (b) Similar figures differ only in scale.

• (c) If two figures are rescaled by the same factor, the ratio of their areas is unchanged.

For in Figure 5.3 we may regard triangle ABD as described on side $\overline{AB}$ of triangle ABC, triangle ACD as described on side $\overline{AC}$, and triangle ABC as described on its own hypotenuse $\overline{\mathit{BC}}$. (Alternatively, we may consider Figure 5.4, in which the triangles ABD, ACD, and ABC of Figure 5.3 have been reflected around the lines $\overline{\mathit{AB}}$, $\overline{\mathit{AC}}$ and $\overline{\mathit{BC}}$ to form the triangles ABD′, ACD″, and A′BC.⁵) Triangles ABD, ACD, and ABC are similar because their corresponding angles are equal, and since triangles ABD and ACD exactly fill up triangle ABC,

$$\displaystyle{\text{area of}\ \mathit{ABD} + \text{area of}\ \mathit{ACD} = \text{area of}\ \mathit{ABC}.}$$

The proposition therefore holds for those triangles. But the general case then follows from (b) and (c), since F ₁, F ₂, and F ₃ are similar by hypothesis and are scaled, like triangles ABD, ACD, and ABC, in proportion to the lengths of $\overline{\mathit{AB}}$, $\overline{\mathit{AC}}$ and $\overline{\mathit{AD}}$. So, for some constant k, we have

$$\displaystyle{ \frac{\text{area of}\ F_{1}} {\text{area of}\ \mathit{ABD}} = \frac{\text{area of}\ F_{2}} {\text{area of}\ \mathit{ACD}} = \frac{\text{area of}\ F_{3}} {\text{area of}\ \mathit{ABC}} = k.}$$

Multiplying the first displayed equation above by k then gives

Fig. 5.4

Figure 5.3 “unfolded”

$$\displaystyle{\text{area of}\ F_{1} + \text{area of}\ F_{2} = \text{area of}\ F_{3},\qquad \qquad \qquad \qquad \qquad \qquad \mathrm{q.e.d.}}$$

The generality of Proposition VI,31, coupled with the economy of means used to prove it, is breathtaking. So why did Euclid give the less general Proposition I,47 with its more involved proof? Presumably, as Heath says (Heath 1956, vol. I, p. 355), because in the plan of exposition that Euclid adopted for the Elements, the development of Eudoxus’s ingenious theory of proportions (needed in order for similarity theory to apply to incommensurable as well as commensurable magnitudes) was postponed to Book V, whereas the more restricted form of the Pythagorean Theorem given in I,47 was needed early on. Alternatively, as Heath also suggests, it may be that a proof of the Pythagorean Theorem based on similarity theory was first advanced prior to Eudoxus’s work, but was recognized to apply only to commensurable magnitudes, so that an alternative proof independent of similarity theory (such as that given for Proposition I,47) was desired and subsequently found.

Note that, despite its reliance on similarity theory, the proof of Proposition VI,31 also involves a comparison of areas, in contrast to the next proof (one commonly given today):

• 3. The Pythagorean Theorem: Given any right triangle whose hypotenuse $a = \vert \overline{BC}\vert$ , let $b = \vert \overline{AC}\vert$ and $c = \vert \overline{AB}\vert$ . Then $a^{2} = b^{2} + c^{2}$ .

FormalPara Proof:

In Figure 5.3, let $y = \vert \overline{\mathit{BD}}\vert$. Triangle ABC is similar both to triangle ABD and to triangle ACD (because corresponding angles are equal), so their corresponding sides are proportional. In particular, $a/c = c/y$ and $a/b = b/(a - y)$. That is, ay = c ² and $a(a - y) = a^{2} -\mathit{ay} = b^{2}$. So $a^{2} - c^{2} = b^{2}$.

The proof above is based on the same diagram used to prove Proposition VI,31. But neither the statement of the Pythagorean Theorem in 3. (the form in which it is usually stated), nor the proof just given, makes any reference to areas; only length relationships are involved. Thus both the meaning of the proposition stated in 3. and the argument used to justify it are conceptually distinct from Euclid’s propositions I,47 and VI,31. Nonetheless, it is worth noting in passing the connection between Figure 5.3 and the bottom part of Euclid’s ‘windmill’ diagram (Figure 5.1). For if we modify Figure 5.3 by erecting the square BCFE on $\overline{\mathit{BC}}$ and extending $\overline{\mathit{AD}}$ to meet the opposite side of that square at G (Figure 5.5), then the rectangle BDGE has height a and width y, and the first proportion used in the proof of 3. shows that $y = c^{2}/a$; so BDGE has area c ². Likewise, rectangle DCFG has area b ², since it has height a and width $a - y = b^{2}/a$.

Fig. 5.5

The base of the ‘windmill’ diagram

The proof given for 3. is attractive from a pedagogical standpoint: It is concise, the diagram is much simpler than that for Euclid’s proof of I,47, and the argument is easy for students to follow (much more so than that for VI,31, because both the idea of a special case implying the truth of a more general statement and the recognition that the triangles in Figure 5.3 are an instance of the configuration described in the statement of VI,31 are difficult for novices to grasp). However, even after being given Figure 5.3 and told to invoke properties of similar triangles, students may have difficulty discovering that proof, since they may have trouble finding the right set of proportional relations to use (a difficulty that may nonetheless make the exercise a valuable one for improving students’ appreciation of the effort and creativity involved in finding proofs).

• 4. The trigonometric form of the Pythagorean Theorem: Let θ be an acute angle in a right triangle. Then $\cos ^{2}\theta +\sin ^{2}\theta = 1$.

FormalPara Proof:

Since ratios of sides are unaffected by scaling, it suffices to consider a right triangle ABC with hypotenuse of length 1 and right angle at A. Let angle θ be at vertex C, place it (for convenience) in standard position on a rectangular coordinate system, and erect the altitude $\vert \overline{\mathit{AD}}\vert$ (Figure 5.6). In triangle ACD, $\cos \theta = \vert \overline{\mathit{CD}}\vert /\vert \overline{\mathit{AC}}\vert = \vert \overline{\mathit{CD}}\vert /\cos \theta$, and in triangle BAD, $\sin \theta = \vert \overline{\mathit{BD}}\vert /\vert \overline{\mathit{AB}}\vert = \vert \overline{\mathit{BD}}\vert /\sin \theta$, since angle BAD = θ. So $1 = \vert \overline{\mathit{CD}}\vert + \vert \overline{\mathit{BD}}\vert =\cos ^{2}\theta +\sin ^{2}\theta$.

Fig. 5.6

$\cos ^{2}\theta +\sin ^{2}\theta = 1$

Apart from its orientation and scaling, Figure 5.6 is the same as Figure 5.3, and if the labels a, b, c, and y are defined as in 3., then $\cos \theta = b/a$ and $\sin \theta = c/a$, whence ay = c ² and $a(a - y) = b^{2}$, as in the earlier proof. Should the proof of 4. then be regarded merely as a variant of the proof of 3.?

The situation is similar to the relation between the gnomon and induction proofs considered in Chapter 3 The trigonometric proof of statement 4. is computationally simpler than the algebraic proof of 3., even though algebraically equivalent to it; and the geometric representation of the expression cos² θ + sin² θ as the length of the hypotenuse adds perspicuity to the proof of 4. as well. Conceptually, then, the two arguments are distinct: an example of how a judicious choice of primitive concepts (here, ratios of lengths rather than lengths themselves) can make a proof both easier to carry out and easier to understand and remember.

5.1 Two dissection proofs

Dissection proofs, used to show the equality of areas, are of two kinds:

Either:

(i) It is shown that two geometric figures of different shapes can be decomposed into the same set of non-overlapping pieces, differently arranged. (Such figures are said to be equidecomposable.)

or:

(ii) It is shown that the same figure can be decomposed into two different sets of non-overlapping pieces, each set containing some pieces congruent to pieces in the other, so that equal areas are left after removal of different congruent pieces.

Such proofs have surprisingly wide applicability, in view of

The Bolyai-Gerwin Theorem: Any two polygons of equal area are equidecomposable.⁶

The two proofs that follow are of the second type.

• 5. A proof without words (See Figure 5.7.)

  Fig. 5.7

  

  A simple dissection proof

This ancient dissection proof compels immediate assent, even from young students with no algebraic background, that the area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of the squares on the other two sides. To understand it, all that is required is knowledge that the angles opposite the legs in a right triangle are complementary (in order to confirm that the sides of the large square in the left diagram are straight line segments).

The diagram on the right is in effect a geometric representation of the identity $(a + b)^{2} = a^{2} + b^{2} + 2\mathit{ab}$, so given the latter, only the diagram on the left is needed to carry out the proof. On the other hand, with a slight alteration, the two diagrams in Figure 5.7 may be adapted to serve a quite different purpose: that of showing that $\sin (\alpha +\beta ) =\sin \alpha \cos \beta +\sin \beta \cos \alpha$ (See Figure 5.8 below, taken from Nelsen 2000, p. 40, where the figure is credited to Volker Priebe and Edgar A. Ramos. ©The Mathematical Association of America 2013. All rights reserved.)

Fig. 5.8

Sine of the sum of two angles

A variation of dissection proof 5., due to the twelfth-century Indian mathematician Bhāskara, is also well-known, and a diagram much like that on the right side of Figure 5.7 is used to illustrate proposition II,4 of Euclid’s Elements. So the question again arises, could not Euclid have proved proposition I,47 more simply by using those diagrams? Heath thought that the only objection to that idea was that such dissection proofs had “no specifically Greek” character (Heath 1956, vol. I, p. 355). Knorr, however, considered that objection “unjust” (Knorr 1975, p. 178 and fn 18 thereto, pp. 204–5). The idea that the passage from the left to the right diagram in Figure 5.7 requires spatial translation of the constituent triangles, an operation not justified by Euclid’s axioms, also does not hold up to scrutiny, for two reasons. First, Euclid’s proof of I,47 relies on proposition I,4 (the side-angle-side criterion for congruence), whose proof, as noted earlier, itself involves spatial displacement of a figure. Second, the two diagrams in Figure 5.7 can be superimposed in a single diagram that is constructible by Euclidean methods from a given right triangle ABC. (See the remark following the next proof.)

6. A proof involving congruent pentagons: Given right triangle ABC with right angle at C, construct squares on sides $\overline{\mathit{BC}},\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$ and label them 2, 3, and 4, respectively. (See Figure 5.9.) Extend the side of square 2 opposite $\overline{\mathit{BC}}$ and the side of square 3 opposite $\overline{\mathit{AC}}$ until they meet, and draw the diagonal from C to that intersection point. Label the resulting triangles 5 and 6. Label the vertices of square 4 diagonally opposite to A and B as D and E, respectively, and draw the perpendicular from D to the extension of $\overline{\mathit{CB}}$ and the perpendicular from E to the extension of $\overline{\mathit{CA}}$. Label the resulting triangles on $\overline{\mathit{AE}}$ and $\overline{\mathit{BD}}$ as 7 and 8, respectively. Regions 1, 2, 3, 5, and 6 together form a pentagon that is congruent to the pentagon formed by regions 1, 4, 7, and 8; and since triangles 5, 6, 7, and 8 are each congruent to triangle ABC, if triangles 1, 5, and 6 are removed from the first of those pentagons and triangles 1, 7, and 8 from the second, the remaining areas must be equal. That is, the area of square 4 equals the sum of the areas of squares 2 and 3. q.e.d.

Fig. 5.9

Overlapping congruent pentagons

Note that in both proofs 5. and 6. triangles congruent to the original are constructed on the sides of the squares in question. But there are only five triangles in Figure 5.9, while there are eight in the two diagrams of Figure 5.7.

As noted earlier, it is possible to combine the two diagrams of Figure 5.7 into one figure and then carry out the dissection. In particular, if two further triangles congruent to ABC are added to Figure 5.9, one (9) below $\overline{\mathit{AB}}$ oriented so as to form a rectangle with ABC and the other (10) in the same orientation below $\overline{\mathit{DE}}$, the resulting Figure 5.10 may be viewed as the two squares of Figure 5.7 placed so as to overlap in triangle ABC.

Fig. 5.10

The two parts of Figure 5.7 combined

The last of the proofs to be considered here is another ‘proof without words’ (the forty-first of those on Bogomolny’s site, credited there to Geoffrey Margrave of Lucent Technologies) which, like proof 5., requires only knowing that the angles opposite the legs in a right triangle are complementary. Unlike the dissection proof, however, it is based on scaling (an operation performable by Euclidean means according to proposition VI,12 of the Elements). Like the ‘folding bag’ proof in Figure 5.4, three triangles each similar to the original are employed, but the conclusion results not from overlap of their areas, but from the equality of the lengths of opposite sides of a rectangle.

7. Proof by scaling: Given right triangle ABC with hypotenuse of length c and legs of lengths a and b, make three copies of it, scaled, respectively, by the factors a, b, and c, and assemble them to form a rectangle as in Figure 5.11.

Fig. 5.11

Three scaled copies fitted together

Readers may judge for themselves which, if any, of the seven proofs above is simplest or most perspicuous.

5.2 Further consequences and extensions

Euclid’s Proposition VI,31 extended Proposition I,47 to ‘arbitrary’ similar figures described on the sides of a right triangle. In another direction, the Law of Cosines provides an extension of I,47 to arbitrary triangles. It includes the Pythagorean Theorem as a special case and also implies its converse.

In Euclid’s treatment, the converse is established in Proposition I,48 (the last in Book I) as an easy consequence of I,47. Specifically (see Figure 5.12), given a triangle ABC in which the square on one of the sides (say $\overline{\mathit{BC}}$) is equal to the sum of the squares on the other two sides, construct a segment perpendicular to $\overline{\mathit{AC}}$ at A of length $c = \vert \overline{\mathit{AB}}\vert$ and join its terminal point D to C. Letting $a = \vert \overline{\mathit{BC}}\vert,b = \vert \overline{\mathit{AC}}\vert$, and $d = \vert \overline{\mathit{CD}}\vert$, then $a^{2} = b^{2} + c^{2}$ by assumption, and $b^{2} + c^{2} = d^{2}$ by I,47. So a = d and triangle ACD is congruent to triangle ABC by the side-side-side criterion (I,8). Hence angle BAC is right.

Fig. 5.12

The converse of the Pythagorean Theorem

The Law of Cosines is also a consequence of the Pythagorean Theorem. Today, after extending the definition of the trigonometric functions to the interval [0, π∕2], the proof is usually carried out by applying the distance formula to a triangle positioned as in Figure 5.13 (which illustrates the obtuse-angled case). The Law of Cosines in its trigonometric form of course does not appear in the Elements. But geometric equivalents for the obtuse- and acute-angled cases are stated and proved as propositions 12 and 13 in Book II. Their statements are:

Fig. 5.13

The Law of Cosines derived from the Pythagorean Theorem

Proposition II, 12:

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the [sum of the] squares on the sides containing the obtuse angle by twice the rectangle containing one of the sides, namely that on [the extension of which] the perpendicular falls, and the straight line cut off [from that extension] outside [the triangle] by [that] perpendicular. (Figure 5.14(a))

Fig. 5.14

Euclid’s propositions II, 12 and II,13

Proposition II, 13:

In triangles [containing an acute angle], the square on the side subtending [that] acute angle is less than the [sum of the] squares on the sides containing [that] angle by twice the rectangle contained by one of the sides about [that] acute angle, namely that on which the perpendicular falls, and the straight line cut off [on that side] within [the triangle] by the perpendicular towards [that] acute angle. (Figure 5.14(b))

Euclid’s proofs of those propositions both rely on the Pythagorean Theorem, together with geometric analogs (Propositions II,4 and II,7) of the algebraic identities $(a + b)^{2} = a^{2} + b^{2} + 2\mathit{ab}$ and $(a - b)^{2} = a^{2} + b^{2} - 2\mathit{ab}$. However, in his commentaries on Propositions II, 12 and II,13, Heath shows that each of those results can alternatively be proved in the same manner as I,47, using variants of the ‘windmill’ diagram (Heath 1956, vol. I, pp. 404–405 and 407–408. For Proposition II,13 there are three cases to consider, depending on whether one of the other angles is right, one is obtuse, or all are acute.) Thus, for triangles not containing a right angle, the Law of Cosines can be proved independently of, but by the same method as, Euclid’s first proof of the Pythagorean Theorem.

Very recently, a uniform proof of the Law of Cosines, valid for all angles and independent of the Pythagorean Theorem, has been given by John Molokach (http://www.cut-the-knot.org/pythagoras/CosLawMolokach.shtml). It is presented here as a final example of a simple, conceptually distinct proof of a statement that implies the Pythagorean Theorem.

8. Direct derivation of the Law of Cosines: Given any triangle ABC, let a, b, and c denote the sides opposite angles A, B, and C, respectively. At least two of the angles, say A and B, must be acute. Then (see Figure 5.15), regardless of whether angle C is acute, right, or obtuse:

Fig. 5.15

(a) The Law of Cosines: all angles acute. (b) The Law of Cosines: angle C right. (c) The Law of Cosines: angle C obtuse

$$\displaystyle{ a = b\cos C + c\cos B, }$$

(1)

$$\displaystyle{ b = a\cos C + c\cos A, }$$

(2)

$$\displaystyle{ c = a\cos B + b\cos A. }$$

(3)

Multiplying (1) by a, (2) by b and (3) by c then gives

$$\displaystyle{ a^{2 } = \mathit{ab}\cos C + \mathit{ac}\cos B, }$$

(4)

$$\displaystyle{ b^{2 } = \mathit{ab}\cos C + \mathit{bc}\cos A, }$$

(5)

$$\displaystyle{ c^{2 } = \mathit{ac}\cos B + \mathit{bc}\cos A. }$$

(6)

Subtracting any two of these equations from the third then gives one of the three forms of the Law of Cosines; e.g., subtracting (4) and (5) from (6) yields that $c^{2} - a^{2} - b^{2} = -2\mathit{ab}\cos C$. q.e.d.