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Home , Cyclic quadrilateral, Parallelogram, Quadrilateral

WILLIAM DUNHAM Quadrilaterally Speaking

mong the figures of plane , the holds to represent both the segment from A to C and the length of a special place. are simple, basic, and this segment. Which is meant should be clear from the con- A unembellished, yet their geometric importance can- text.) not be overemphasized. All have three medians, three alti- We shall examine a trio of relating the sides and tudes, and one , and all possess both inscribed and of . Although far from new, they seem circumscribed . When two or more triangles get to- to be relatively unknown, even among the mathematically gether, they can be the spitting image of one another (i.e., be preoccupied. Subscribing to the adage that good things bear congruent) or at least bear a strong family resemblance (i.e., repeating, we consider them in turn. be similar). And, as everyone knows, congruent and similar triangles are critical to the logical development of geometry. But triangles have one major shortcoming when compared ’s to their polygonal cousins: they lack diagonals. After draw- ing a triangle’s three sides, the mathematician finds that no This result appears in Book I of the Almagest, the astronomi- unconnected vertices remain, and consequently the “diago- cal masterpiece of Claudius Ptolemy (ca. 85–165). A word of nal” of a triangle is the geometric counterpart of one hand warning: the likeness we have provided of Ptolemy is not to clapping. be trusted. Indeed, it looks suspiciously like the images we In order to explore the interconnection among sides and have of Homer, Aeschylus, and Socrates—not to mention diagonals, it is necessary to step up to quadrilaterals. For the those of Noah, Nebuchadnezzar, and Neptune. Portraits of purposes of this article, all quadrilaterals will be convex, with classical males tend to be interchangeable, with everybody both diagonals falling neatly inside. As denoted in Figure 1, more or less resembling Moses, or perhaps Charlton Heston. ABCD will have sides of length a, b, c, and d But Claudius Ptolemy does hold one legitimate distinction: and diagonals of length x = AC and y = BD. (We will use AC he is the earliest major mathematician who possessed both a first and a last name. His illustrious predecessors—from

D

d y

A c

a x

BC b Figure 1

WILLIAM DUNHAM is Truman Koehler Professor of Mathematics at Muhlenberg College. He is the author of Euler: Master of Us All, Journey Through Genius: The Great Theorems of Mathematics, and The Mathematical Universe. Claudius Ptolemy

12 Math Horizons February 2000 © Mathematical Association of America A

D G 1 d y A x a E B a c 1 O a

B b C

Figure 2 Figure 3 to to Apollonius—got by quite well with one Moreover, a famous identity follows as an easy corollary. To see name. (Whether “Euclid” was his first name or his last name it, we first observe that if 2α is the measure of central is a matter that concerns only chronic insomniacs.) AOB in the unit (see Figure 3) and if we draw OG perpen- In any case, before stating Ptolemy’s theorem, we need dicular to AB, then sin α = AG/OA = AG, and so the Definition: A is one that can be inscribed length of chord AB is 2(AG) = 2sinα. in a circle. Now for the identity. Suppose we begin with a pair of having measures 2α and 2β, as shown in Figure 4. Put another way, a quadrilateral is cyclic if a circle can be Place these as central angles within a unit circle, one on circumscribed about it. It should be clear that this is a very either side of BD, and connect points A, B, C, and restrictive condition. For, suppose we refer to the quadrilat- D to generate a cyclic quadrilateral and its diagonals. By our eral in Figure 1 and circumscribe a circle about ∆ABC . Then previous observation, quadrilateral ABCD is cyclic if and only if the D falls exactly upon the circle just drawn— an unlikely circumstance AB = 2sinα, BC = 2sinβ, and AC = 2sin(α + β). to be sure. Furthermore, because DAB and DCB are inscribed in a In spite of the fact that cyclic quadrilaterals are far from semicircle, the corresponding triangles ∆DAB and ∆DCB general, Ptolemy proved a beautiful result about them, namely: are right, and so the implies that

Theorem 1. If ABCD is a cyclic quadrilateral, then the product AD=−=−α=−α=α BD22 AB 44sin 2 21sin 2 2cos of the diagonals is the sum of the products of the opposite sides. and similarly that CD = 2cos β. That is, in Figure 2, ac + bd = xy. Proof. To begin his elegant argument, Ptolemy constructed ∠ABE congruent to ∠DBC, where E lies on AC. A Because ∠BAC and ∠BDC intercept the same arc of the circle, they are congruent. Therefore ∆ ABE is similar to ∆DBC, from which it follows that AB/AE = DB/DC. That is, a/AE = y/c, or equivalently ac = (AE)y (1) Likewise, ∆ ABD is similar to ∆EBC because ∠ADB is con- 2a 1 B D gruent to ∠ECB (they intercept the same arc) and ∠ABD = 2b O ∠ABE + ∠EBD = ∠DBC + ∠EBD = ∠EBC. Thus BD/AD = BC/EC. That is, y/d = b/EC, and so bd = (EC)y. (2) Now simply add equations (1) and (2) to get ac + bd = (AE + EC)y = (AC)y = xy, C and the proposition is proved. QED Ptolemy used this theorem in the Almagest to generate his “Table of Chords,” the precursor of our trigonometric tables. Figure 4

© Mathematical Association of America Math Horizons February 2000 13 A b D x y

a a

q pq–

B b C Figure 5

Thus when we apply Ptolemy’s theorem to ABCD, we get (2sinα)(2cosβ) + (2sin β)(2cosα) = 2[2sin(α + β)], which reduces to sin(α + β) = sinα cos β + cosα sin β, the justly famous identity from trigonometry. Claudius Ptolemy was surely on to something. Whereas Ptolemy’s theorem is restricted to cyclic quadri- laterals, our second is specific to . Yet, as we shall see, it points the way towards the general case.

Theorem 2. If ABCD is a as in Figure 5, then 2a2 + 2b2 = x2 + y2. In the case when ABCD is a with AC = BD = x, Recalling that the opposite sides of a parallelogram are then theorems 1 and 2 reduce (respectively) to congruent, we can restate this as: the sum of the of 2 2 2 2 2 2 the four sides of a parallelogram (i.e., a2 + b2 + a2 + b2) equals aa + bb = x and 2a + 2b = x + x = 2x . the sum of the squares of the diagonals. In short, when both apply, they jointly collapse into the Pythagorean theorem— a most significant crossroads. Proof. With θ as the measure of ∠ABC—and thus π – θ as Alas, the aforementioned results hold only for special the measure of ∠BCD —apply the to ∆ ABC kinds of quadrilaterals. The big challenge is to prove a theo- and ∆DBC to get rem relating the sides and diagonals of a general convex quad- x2 = a2 + b2 – 2abcosθ and y2 = a2 + b2 – 2abcos(π – θ). rilateral. What this relationship might be, and how to prove Then, because cos(π – θ) = – cosθ, we need only add these it, are far from obvious. equations to reach the desired end. QED Fear not. In a 1748 paper, Leonhard Euler (1707–1783) rose to the challenge [1]. We shall present the theorem—as We now have two theorems at our disposal—one pertain- he did—in stages, building toward the general result after a ing to cyclic quadrilaterals, the other to parallelograms— pair of lemmas. (Note in passing that Euler looks nothing and it is natural to ask under what conditions they are simul- like Moses, although he does sport a natty turban.) taneously applicable. That is, when is a quadrilateral cyclic For the purposes of the remaining proofs, we shall explic- and parallelogramic (if there is such a word)? itly draw our quadrilateral as though it is not a parallelo- A moment’s thought suggests that this happens if and gram. The reader can check that the results become triviali- only if the quadrilateral is a rectangle, and the proof of this ties in the case of parallelograms. double implication is easy. In one direction, we note that a cyclic quadrilateral that is also a parallelogram must satisfy Lemma 1. Given quadrilateral ABCD in Figure 6, complete par- both theorems 1 and 2, and so (with our prior notation) allelogram ABCE and draw DE. Then abcd2222222+++ = ACBDDE + + . 22ab22+ xy22+ xy=+= aa bb = . In words, this says that the sum of the squares of a 22 quadrilateral’s four sides equals the sum of the squares of its 2 2 2 Cross multiplication yields 0 = x – 2xy + y = (x – y) , diagonals plus the of the length of this newly created which implies that x = y. But a parallelogram with equal segment DE. Think of DE as measuring how far ABCD devi- diagonals is a rectangle, so this direction is proved. Con- ates from a parallelogram. versely, a rectangle is already a parallelogram and is obvi- ously cyclic, for a circle with center at the intersection of the Proof. In Figure 6, construct CF to AD with CF = diagonals and with half the length of the diagonal AD = d and draw segments AF, BE, BF, DF, and EF. It can be passes through all four vertices. proved (exercise!) that quadrilaterals ADCF and BDEF are

14 Math Horizons February 2000 © Mathematical Association of America D D y d d

c c A E A x Q

a a P

b BC B C b

Figure 8 d

F and so P—the of AC —must also be the midpoint of BE. But then segment PQ connects the of sides Figure 6 BE and BD in ∆BED, and by similar triangles we conclude that 2(PQ) = DE. Squaring both sides proves the lemma. QED themselves parallelograms. Thus, by theorem 2, At last, we establish the general theorem (see Figure 8). 2c2222+= 2 d AC + DF and 2 BD 2 + 2 DE 2 = BE 22 + DF . Euler stated it as: Consequently, Theorem 3. “In any quadrilateral, the sum of the squares of the 22 2 2 2 2 2 22c+− d AC = DF = 2 BD + 2 DE − BE , four sides equals the sum of the squares of the diagonals plus and so four times the square of the line connecting the midpoints of the 22cd22+= 2 BDDEACBE 2 + 2 2 + 2 − 2 . diagonals.” But from parallelogram ABCE, we have 2a2 + 2b2 = Proof. This is now trivial, for we need only combine lemma AC2 + BE2. Adding these last two equations yields 1 and lemma 2: 2222222 22222abcd+++ = ACBDDE + 2 + 2, abcd222+++ 2 = ACBDDE 2 + 2 + 2 and thus a2 + b2 + c2 + d2 = AC2 + BD2 + DE2. QED =++AC22 BD4 PQ 2 22 2 As desired, this lemma relates the four sides and two di- =++xy4. PQ QED agonals of a general quadrilateral. But it is less than optimal, Notice how the point E, which Euler introduced some- requiring as it does the “extraneous” segment DE. To tidy up what artificially at the outset, has disappeared from the final this defect, Euler pushed on: theorem—truly, Euler giveth and Euler taketh away. With Lemma 2. For quadrilateral ABCD with completed parallelogram this argument, he had established a curious property of quad- ABCE as above, bisect diagonal AC at P and diagonal BD at Q rilaterals that was, as he put it, “neither enunciated nor proved and draw segment PQ connecting these midpoints (see Figure 7). to this point.” Then DE2 = 4PQ2. Euler noted a pair of interesting consequences: Proof. We first assert that P lies on BE. This follows be- The sum of the squares of the four sides of a convex cause the diagonals of parallelogram ABCE bisect each other, quadrilateral is always greater than or equal to the sum of the squares of its diagonals. D The sum of the squares of the four sides of a convex quadrilateral equals the sum of the squares of its diago- nals if and only if the quadrilateral is a parallelogram. The latter follows because, under the condition of equality, the length of the segment PQ is zero. Consequently, the diagonals A E Q bisect each other, and the quadrilateral is a parallelogram. For the sake of variety, we conclude with an alternate— and shorter—demonstration of theorem 3. Although Euler P needed the extraneous segment DE, this argument involves only the original quadrilateral. And, whereas Euler used syn- BC thetic geometry, this proof rests upon trigonometry, in par- Figure 7 ticular upon a five-fold (!) application of the law of cosines.

© Mathematical Association of America Math Horizons February 2000 15 D and this last is exactly the expression within the square brack- ets of equation (3) above. A final substitution gives us, as d before, v a2 + b2 + c2 + d2 = AC2 + BD2 + 4PQ2. QED c A s Q Of course much more could be said about quadrilaterals. a There is, for instance, the fact that if the sums of the squares O of opposite sides of a quadrilateral are equal (i.e., if a2 + c2 = a 2 2 u P b + d ), then its diagonals must be . Or there t is ’s wonderful formula giving the of a cy- BCclic quadrilateral in terms of the lengths of its four sides. b For now, however, we must leave this topic, reminded once Figure 9 again of the unexpected patterns lurking beneath the surface of . It was Howard Eves [3] who percep- tively observed that this subject “though elementary, is often Theorem 3 Revisited. [2] In any quadrilateral, the sum of the far from easy.” As the preceding theorems reveal, quadrilat- squares of the four sides equals the sum of the squares on the diago- erals can hold their share of surprises. nals plus four times the square of the line connecting the midpoints of the diagonals. Note. I wish to thank Drs. Penny Dunham and Elyn Rykken of Muhlenberg College for their helpful suggestions Proof. As illustrated in Figure 9, we again consider quad- in the preparation of this article. Q rilateral ABCD. With O as the intersection of the diagonals, we let OA = s and OC = t, and we stipulate without loss of generality that s ≤ t. Likewise, let OB = u and OD = v, with References u ≤ v. Finally, we let α be the measure of ∠COD. Of course it 1. Leonhard Euler, Opera Omnia, Ser. 1, Vol. 26, pp. 29–32. follows that α is also the measure of ∠AOB and π – α is the 2. Thanks to Elyn Rykken for showing me this proof. measure of ∠BOC and ∠DOA. 3. Howard Eves, A Survey of Geometry, Allyn and Bacon, Boston, Apply the law of cosines individually to ∆AOB, ∆BOC, 1963, p. 64. ∆COD, and ∆DOA, while remembering that cos(π – α) = – cos α: asu222=+ −2cos suα but222=++2cos utα ctv222=+ −2cos tvα dvs222=++2cos. vsα Upon adding, we have

abcd222+++ 2 (3) =+++−222stuv22 2 22( 2 suuttvvs −+− )cosα =+()()()()2()()cosst2222 ++ uv +− ts +− vu − tsvu − − α

22 22ts−− vu ts −− vu =++AC BD 42cos. + − α 22 22 But recall that P is the midpoint of diagonal AC and Q the midpoint of diagonal BC, as shown in Figure 9 . As a conse- quence, ts+− ts OP=−=−= OC PC t 22 and vu+− vu OQ=−=−= OD QD v . 22 Therefore, the law of cosines applied to ∆OPQ yields PQ222=+− OP OQ2cos() OP( OQ ) α 22 ts−− vu ts −− vu =+ −2cos,α 22 22

16 Math Horizons February 2000 © Mathematical Association of America