MST Topics in History of Mathematics Euclid’s Elements and the Works of Archimedes

Paul Yiu

Department of Mathematics Florida Atlantic University

Summer 2014

June 25 Chapter 0

Greek geometry before Euclid

Chronology

-586 Babylonian captivity -585 Thales of Miletus; deductive geometry -580 Birth of Pythagoras -540 Pythagorean arithmetic and geometry -430 Elements of Hippocrates of Chios -427 Birth of Plato -420 Incommensurables -399 Death of Socrates -360 Eudoxus on proportion and exhaustion -347 Death of Plato -335 Eudemus: History of Geometry -332 Alexandria founded -323 Death of Alexander -322 Death of Aristotle -300 Euclid’s Elements -225 Apollonius: Conics -212 Death of Archimedes +75 Works of Heron of Alexandria +250 Diophantus: Arithmetica +320 Pappus: Mathematical Collections +485 Death of Proclus 2 Greek geometry before Euclid

Thales of Miletus

Theorem (Thales). The angle inscribed in semicircle is a right angle. 3

Other theorems attributed to Thales Eudemus’ History (ca -320):

1. A circle is bisected by a diameter.

2. The base angle of an isosceles triangle are equal. (Euclid I.5)

3. The pairs of vertical angles formed by two intersecting lines are equal. (Eu- clid I.15)

4. If two triangles are such that two angles and a side of one are equal respec- tively to two angles and a side of the other, then the triangles are congruent. (Euclid I.26) 4 Greek geometry before Euclid

Pythaogas of Samos

Proclus, on Euclid I: (p.298)

Eudemus the Peripatetic ascribes to the Pythagoreans the discovery of this theorem, that any triangle has internal angles equal to right angles. He says they proved the theorem in question after this fashion.

A D E

B C Let ABC be a triangle, and through A let DE be drawn parallel to BC. Now since BC, DE are parallel, and the alternative angles are equal, the angle DAB is equal to the angle ABC, and EAC is equal to ACB. Let BAC be added to both. Then the angles DAB, BAC, CAE, that is, the angles DAB, BAE, that is, two right angles, are equal to the three angles of the triangle. Therefore the three angles of the triangle are equal to two right angles. Such is the proof of the Pythagoreans. 5

Iamblichus: On the Pythagorean Life (Thomas, 222–225):

It is related of Hippasus that he was a Pythagoreans, and that, owing to his being the first to publish and describe the sphere from the twelve pentagons, he perished at sea for his impiety, but he received credit for the discovery, though really it all belonged to HIM (for in this way they refer to Pythagoras, and they do not call him by his name). 6 Greek geometry before Euclid

Euclid, Elements X, Scholium i The Pythagoreans were the first to make inquiry into commensurability, having first discovered it as a result of their observation of numbers; for though the unit is a common measure of all numbers they could not find a common measure of all magnitudes. The reason is that all numbers, of whatsoever kind, however they be divided leave some least part which will not suffer further division; but all magnitudes are divisible ad infinitum and do not leave some part which, being the least possible, will not admit of further division, but that remainder can be divided ad infinitum so as to give an infinite number of parts, of which each can be divided ad infinitum; and in sum, magnitude partakes in division of the principle of the infinite, but in its entirety of the principle of the finite, while number in division partakes of the finite, but in its entirety of the infinite. ... There is a legend that the first of the Pythagoreans who made public the investigation of these matters perished in a shipwreck. 7

Democritus

Archimedes, Method:

...incase of those theorems concerning the cone and pyramid of which Eudoxus first discovered the proof, the theorem that the cone is the third part of the cylinder, and the pyramind of the prism, having the same base and equal height, no small share of the credit should be given to Democritus, who was the first to make the assertion with regard to the said figure, though without proof. 8 Greek geometry before Euclid

Duplication of the cube

Theon of Smyrna (Thomas 257):

In his work entitled Platonicus Eratosthenes says that, when the god announced to the Delians by oracle that to get rid of a plague they must construct an altar double of the existing one, their craftsmen fell into great perplexity in trying to find how a solid could be made double of another solid, and they went to ask Plato about it. He told them that the god had given this oracle, not because he wanted an altar of double the size, but he wished, in setting this task before them, to reproach the Greeks for their neglect of mathematics and their contempt for geometry. 9

Eutocius: Commentary on Archimedes’ Sphere and Cylinder, II (Thomas 257–259):

It became a subject of inquiry among geometers in what manner one might double the given solid while it remained the same shape, and this problem was called the duplication of the cube; for given a cube, they sought to double it. When all were for a long time at a loss, Hippocrates of Chios first conceived that, if two mean proportionals could be found in continued proportion between two straight lines, of which the greater was double the lesser, the cube would be doubled, so that the puzzle was by him turned into no less a puzzle. After a time, it is related, certain Delians, when attempting to double a certain altar in accordance with an oracle, fell into the same quandary, and sent over to ask the geometers who were with Plato in the Academy to find what they sought. When these men applied themselves diligently and sought to find two mean proportionals between two given straight lines, Archytas of Taras is said to have found them by the half-cylinders, and Eudoxus by the so-called curved lines; but it turned out that all their solutions were theoretical, and they could not give a practical construction and it it to use, except to a certain small extent Menaechmus, and that with difficulty. An easy mechanical solution, was however, found by me, and by means of it I will find, not only two means to the given straight lines, but as many as may be enjoined. 10 Greek geometry before Euclid

Nicomedes’ solution Given: Two lines AD and DC at right angles. To construct: Two continued mean proportionals between AD and CD.

M

A D

F

B E C K H

T

Z

Construction: Complete the rectangle ADCB and take the midpoints E, F of BC and AB. Join DF to meet CB produced at H. Construct EZ perpendicular to BC so that CZ = AF . Join HZ and construct the parallel through C. On this parallel construct a point T so that ZT meet BC produced at K with TK = CZ. Then AD : AM = AM : CK = CK : CD, and AM, CK are two continued mean proportionals between AD and CD. 11

MAA Focus June/July 2014, 27. x and y are two continuous mean proportionals between 1 and 2:

A

x

1 1 Y

1

B 1 C y X

1

Z 12 Greek geometry before Euclid

Cubic root of 2 by paper folding B. Casselman, If Euclid had been Japanese, Notices of AMS, 54 (2007) 626–628.

A paper square ABCD is divided into three strips of equal area by the parallel lines PQand RS. The square is then folded so that C falls on AD and S falls on PQ (as C in the second√ diagram). AC = 3 2 Then CD .

A B A B

P Q P S

C

R S R

D C D C 13

Trisection of an angle

Archimedes [Book of Lemmas, Proposition 8] Given angle AOB with OA = OB (contained in a circle, center O), construct a line through A such that the intercept between the circle and the line BO has the same length as the radius of the circle. ∠AOC = 1 ∠AOB Then 3 .

A A C

O

B 14 Greek geometry before Euclid

Trisection of an angle To triangle an angle AOB, pass a line through O such that the intercept between the parallel and the perpendicular at A to the line OB is 2 · OA. Then this line is a trisector of angle AOB.

A E

M

D

O B 15

Angle trisection with the use of conics To trisect an angle AOB, construct the hyperbola with focus A, directrix OM, and eccentricity 2, to intersect the arc AB at C. Then ∠AOB =3∠AOC.

C

K

B A M P

O 16 Greek geometry before Euclid

The quadratrix of Hippias A horizontal line HK (with initial position AB) falls vertically, and a radius OP (with initial position OA) rotates about O, both uniformly and arrive at OC at the same time. The locus of the intersection Q = HK ∩ OP is the quadratrix.

A B

P

Q H K

O C 17

Trisection of an angle by the quadratrix

B

P

K Q

A O To trisect angle AOB, let OB intersect the quadratrix at P . Trisect the segment OP at K. Construct the parallel through K to OA to intersect the quadratrix at Q. Then OQ is a trisector of angle AOB. 18 Greek geometry before Euclid

Angle trisection by paper folding B. Casselman, If Euclid had been Japanese, Notices of AMS, 54 (2007) 626–628.

A M B A M B

P  P Q P

R

D R S R

D C D C 19

Quadrature of the circle

Proclus on Euclid I.45: It is my opinion that this proposition is what led the ancients to attempt the squaring of the circle. For if a parallelogram can be found equal to any rectilinear figure, it is worth inquiring whether it is not possible to prove that a rectilinear figure is equal to a circular area. 20 Greek geometry before Euclid

Pappus on the quadratrix For the squaring of the circle a certain line was used by Dionstratus’ and Nicomedes and certain other more recent geometers, and it takes the name from its special property: for it called by them the quadratrix, ....

If ABCD is a square and BED the arc of a circle with center C, while BHT is a quadratrix generated in the aforesaid manner, it is proved that the ratio of the arc DEB towards the straight line BC is the same as that of BC towards the straight line CT.

arc BED : AB = AB : CT.

B A

E

H

C T D Construct a length b such that CT : BC = BC : b. b BC BED Then the rectangle with sides 2 and is equal to the quadrant . 21

Incommensurables

Aristotle, Prior Analytics For all who argue per impossibile infer by syllogism a false conclusion, and prove the original conclusion hypothetically when something impossible follows from a contradictory assumption, as, for example, that the diagonal [of a square] is incommensurable [with the side] because odd numbers are equal to even if it is assumed to be commensurate. It is inferred by syllogism that odd numbers are equal to even, and proved hypothetically that the diagonal is commensurate, since a false conclusion follows from the contradictory assumption. 22 Greek geometry before Euclid

Incommensurability of the diagonal and side of a square If the diagonal d and the side s of a square have a (unit) common measure, these are numbers satisfying d2 =2s2. d2 is an even number. Therefore, d is an even number. Since d and s do not have common measure, s is an odd number. Writing d =2m,wehave(2m)2 =2s2, 4m2 =2s2, and s2 =2m2. This shows that is s2 an even number. Therefore, s is also an even number. But s cannot be both odd and even. This contradiction shows that the diagonal and the side of a square are incommensurable. 23

Euclid’s Elements

There is no royal road to geometry.

Euclid to Ptolemy and Alexander 24 Greek geometry before Euclid

Proclus on Euclid’s Elements

It is a difficult task in any science to select and arrange properly the elements out of which all others matters are produced and into which they can be resolved. Of those who have attempted it some have brought together more theorems, some less; some have used rather short demonstrations, others have extended their treatment to great lengths; some have avoided the reduction to impossibility, others proportion; some have devised defenses in advance against attacks upon the starting points; and in general many ways of constructing elementary expositions have been individually invented. Such a treatise ought to be free of everything superfluous, for that is a hindrance to learning; the selections chosen must all be coherent and conducive to the end proposed, in order to be of the greatest usefulness for knowledge; it must devote great attention both to clarity and to conciseness, for what lacks these qualities confuses out understanding; it ought to aim at the comprehension of its theorems in a general form, for dividing one’s subject too minutely and teaching it by bits makes knowledge of it difficult to attain. Judged by all these criteria, you will find Euclid’s introduction superior to others. 25

Euclid’s Elements Books Subject I – VI Plane geometry VII – IX Number theory X Theory of irrational constructible quantities XI–XIII Solid geometry

Book I II III IV V VI Total Definitions 23 2 10 7 18 5 65 Common notions 5 Postulates 5 Propositions 48 14 37 16 25 33 173

Book VII VIII IX X XI XII XIII Total Definitions 22 16 28 66 Propositions 39 27 36 115 39 18 18 292 26 Greek geometry before Euclid

The first definitions

Definitions. (I.1). A point is that which has no part. (I.2). A line is breadthless length. (I.3). The extremities of a line are points. (I.4). A straight line is a line which lies evenly with the points on itself.

Definition (I.10). When a straight line set up on a straight line makes the adjacent angles equal to one an- other, each of the equal angles is [a] right [angle], and the straight line standing on the other is called a perpendicular to that on which it stands.

Definition (I.23). Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. 27

The common notions

(1) Things which equal the same thing also equal one another. a = b and b = c =⇒ a = c.

(2) If equals are added to equals, then the wholes are equal. a = b and c = d =⇒ a + c = b + d.

(3) If equals are subtracted from equals, then the remainders are equal. a = b and c = d =⇒ c − a = d − b.

(4) Things which coincide with one another equal one another.

(5) The whole is greater than the part. 28 Greek geometry before Euclid

The postulates

Postulate 1. To draw a straight line from any point to any point. Postulate 2. To produce a finite straight line continuously in a straight line. Postulate 3. To describe a circle with any center and distance. Postulate 4. That all right angles are equal to each other.

Postulate 5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. Chapter 1

Euclid’s Elements, Book I (constructions) 2 Euclid’s Elements, Book I (constructions)

1.1 The use of ruler and compass

Euclid’s Elements can be read as a book on how to construct certain geometric figures efficiently and accurately using ruler and compass, and ascertaining the validity. The first three postulates before Book I are on the basic use of the ruler and the compass.

Postulate 1. To draw a straight line from any point to any point. With a ruler (straightedge) one connects two given points A and B to form the line (segment) AB, and there is only one such line. This uniqueness is assumed, for example, in the proof of I.4.

Postulate 2. To produce a finite straight line continuously in a straight line. Given two points A and B, with the use of a ruler one can construct a point C so that the line (segment) AC contains the point B. The first two postulates can be combined into a single one: through two distinct points there is a unique straight line. 1.1 The use of ruler and compass 3

Postulate 3. To describe a circle with any center and distance. This distance is given by a finite line (segment) from the center A to another point B. With the use of a collapsible compass, one constructs a circle with given center A to pass through B. We denote this circle by C(A, B). Euclid I.2 shows how to construct a circle with a given center and radius equal to a given line (seg- ment). 4 Euclid’s Elements, Book I (constructions)

Definition (I.20). Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.

Remark. Euclid seems to take isosceles and scalene in the exclusive sense. But it is more convenient to take these in the inclusive sense. An isosceles triangle is one with two equal sides, so that an equilateral triangle is also isosceles. 1.1 The use of ruler and compass 5

Euclid (I.1). On a given finite straight line to construct an equilateral triangle.

C

D E A B

Given: Points A and B. To construct: Equilateral triangle ABC. Construction: Construct the circles C(A, B) and C(B,A) to intersect at a point C. ABC is an equilateral triangle. 6 Euclid’s Elements, Book I (constructions)

The second proposition is on the use of a collapsible compass to transfer a seg- ment to a given endpoint.

Euclid (I.2). To place a straight line equal to a given straight line with one end at a given point.

D A

C B L

G

Given: Point A and line BC. To construct: Line AL equal to BC. Construction: (1) An equilateral triangle ABD, [I.1] (2) the circle C(B,C). (3) Extend DB to intersect the circle at G. (4) Construct the circle C(D, G) and (5) extend DA to intersect this circle at L. AL = DL − DA = DG − DB = BG = BC. Therefore the circle C(A, BC) can be constructed using a collapsible compass. 1.1 The use of ruler and compass 7

Euclid (I.3). Given two unequal straight lines, to cut off from the greater a straight line equal to the less. 8 Euclid’s Elements, Book I (constructions)

Euclid (I.9). To bisect a given rectilineal angle.

A

D E

F BC Given: Angle BAC. To construct: Line AF bisecting angle ABC. Construction: (1) Choose an arbitrary point D on AB. (2) Construct E on AC such that AD = AE. [I.3] (3) Construct an equilateral triangle DEF on DE (so that F and A are on opposite sides of DE). [I.1] The line AF is the bisector of the angle BAC. 1.1 The use of ruler and compass 9

Euclid (I.10). To bisect a given finite straight line.

C

A D B Given: Line segment AB. To construct: The midpoint of AB. Construction: (1) An equilateral triangle ABC, [I.1] (2) the bisector of angle ACB [I.9] (3) to meet AB at D. AD = DB. 10 Euclid’s Elements, Book I (constructions)

1.2 Perpendicular lines

Definition (I.10). When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.

Postulate 4. That all right angles are equal to each other.

Euclid (I.11). To draw a straight line at right angles to a given straight line from a given point on it.

F

AB D C E Given: A straight line AB and a point C on it. To construct: A line CF perpendicular to AB. Construction: (1) Take an arbitrary point D on AC, and construct E on CB such that DC = CE. (2) Construct an equilateral triangle DEF. [I.1] The line CF is perpendicular to AB. Proof : In triangles DCF and ECF, DC = EC, by construction CF = CF, DF = EF sides of equilateral triangle DCF ≡ ECF SSS ∠DCF = ∠ECF CF ⊥ AB. 1.2 Perpendicular lines 11

Euclid (I.12). To draw a straight line perpendicular to a given infinite straight line from a given point not on it.

C

AB G H E D

Given: A straight line AB and a point C not on the line. To construct: A line CH perpendicular to AB. Construction: (1) Take a point D on the other side of the line and construct the circle C(C, D) to meet the line at E and G. Bisect the segment EG at H [I.10] CH is perpendicular to AB. Proof : Triangles ECH and GCH are congruent. [SSS] ∠ECH = ∠GCH. CH ⊥ AB. 12 Euclid’s Elements, Book I (constructions)

1.3 Isosceles triangles

Euclid (I.5). In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.

A

B C

F G

DE

Given: Triangle ABC with AB = AC. To prove: ∠ABC = ∠ACB. Construction: Extend AB and AC to D and E respectively. Choose an arbitrary point F on BD, and construct G on CE such that CG = BF. [I.3] Proof : In triangles CAF and BAG, CA = BA, isosceles triangle ∠CAF = ∠BAG, AF = AB + BF = AC + CG = AG. ∴ BAG ≡CAF, SAS CF = BG, and ∠BFC = ∠AF C = ∠AGB = ∠CGB. In triangles BFC and CGB, BC = CB, ∠BFC = ∠CGB, CF = BG, proved above ∴ BFC ≡CGB SAS ∠BCF = ∠CBG ∠ABC = ∠ABG − ∠CBG = ∠CAF − ∠BCF = ∠ACB. Q.E.D.

Q.E.D.: quod erat demonstrandum, which was to be demonstrated. 1.3 Isosceles triangles 13

This proposition is often called the “bridge of asses” (pons asinorum). According to D. E. Smith (The Teaching of Geometry, 1911, p.174), “[i]t is usually stated that it came from the fact that fools could not cross this bridge, and it is a fact that in the Middle Ages this was often the limit of the student’s progress in geometry. It has however been suggested that the name came from Euclid’s figure, which resembles the simplest type of a wooden truss bridge”. 14 Euclid’s Elements, Book I (constructions)

According to Proclus, Pappus (3rd century A.D.) proved the proposition in a very simple way by considering an isosceles triangle ABC as two triangles, ABC and ACB. From AB = AC, ∠BAC = ∠CAB, AC = AB, it follows from I.4 that ABC ≡ACB, and so ∠ABC = ∠ACB. 1.3 Isosceles triangles 15

Here is another apparently easy proof of I.5: let the bisector of angle BAC meet BC at D. Then ABD ≡ACD by I.4 again. From this we conclude that ∠ABD = ∠ACD, equivalently, ∠ABC = ∠ACB. However, the existence of an angle bisector is justified only in I.9. Euclid is a true constructivist. He does not consider squares before I.46. 16 Euclid’s Elements, Book I (constructions)

Euclid (I.6). If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

Given: Triangle ABC with ∠ABC = ∠ACB. To prove: AB = AC. Proof by contradiction: (1) Suppose AB is greater than AC. Choose a point D on AB such that DB = AC. [I.3] In triangles DBC and ACB, DB = AC, ∠DBC = ∠ACB, BC = CB. ∴ DBC ≡ ACB. SAS This is contradiction since triangle DBC is part of triangle ABC. (2) Suppose AB is less than AC. A similar reasoning also leads to a contradiction. (3) Therefore, AB = AC. Q.E.D. 1.4 Tests for congruence of triangles I.4,8,26 17

1.4 Tests for congruence of triangles I.4,8,26

Euclid did not use the term congruence of triangles. When he says two triangles are equal, he means they are equal in area.

Euclid (I.4). [SAS] If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remain- ing angles equal the remaining angles respectively, namely those which the equal sides subtend.

It is convenient to identify congruent triangles with corresponding vertices and sides. Thus, two triangles ABC and XY Z are congruent if the corresponding sides are equal, namely, BC = YZ, CA= ZX, XY = AB, and the corresponding angles are equal, namely, ∠BAC = ∠YXZ, ∠CBA = ∠ZY X, ∠ACB = ∠XZY.

Euclid I.4 is the first of four valid tests for congruence of triangles. We refer to it as the SAS test. Book I contains two more tests. 18 Euclid’s Elements, Book I (constructions)

Euclid (I.8). [SSS] If two triangles have the two sides equal to two sides respec- tively, and have also the base equal to the base, they will have the angles equal which are contained by the equal straight lines.

Euclid (I.26). [ASA, AAS] If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the re- maining angle. 1.4 Tests for congruence of triangles I.4,8,26 19

The following proposition, though not in Euclid’s Elements, completes the list of congruence tests.

Proposition (RHS). Two right angled triangles with equal hypotenuses and one pair of equal sides are congruent.

This is, of course, is a corollary of I.48. It can, however, be established without invoking this. Let ABC and ABC be two triangles in which the angles ACB and ACB are right angles, AB = AB and AB = AB. On the side of AC opposite to B, construct a triangle ADC congruent to ABC (with AD = AB and CD = CB). Since the angles ACB and ACD are both right angles, B, C, D are a line. Since AB = AD, by I.5, ∠ADC = ∠ABC. Therefore, triangles ABC and ADC are congruent (I.4); so are ABC and ABC. 20 Euclid’s Elements, Book I (constructions)

1.5 Two constructions

Euclid (I.22). Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.

Euclid (I.23). [Copying an angle along a line] On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectlineal angle.

Given: A line AB and an angle DCE. To construct: BAF equal to angle DCE. Construction: Choose arbitrary points D, E on the sides of the angle C. Construct triangle AF G such that CD = AF , CE = AG and DE = FG. [I.22] 1.6 Parallel lines 21

1.6 Parallel lines

Definition (I.23). Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.

Postulate 5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

This famous fifth postulate has a long history, eventually leading to the dis- covery of noneuclidean geometry in the 19th century. Here, we simply remark that it provides an explicit criterion for the intersection of finite lines upon extension. Heath [I.201] pointed out that while Postulate 4 can be proved using other axioms, “[i]t was essential from Euclid’s point of view that it should come before Post. 5, since the condition in the latter that a certain pair of angles are together less than two right angles would be useless unless it were first made clear that right angles are angles of determinate and invariable magnitude”. 22 Euclid’s Elements, Book I (constructions)

Euclid (I.27). If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. (I.28). If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. (I.29). A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. (I.30). Straight lines parallel to the same straight line are also parallel to one an- other. 1.6 Parallel lines 23

Euclid (I.31). Through a given point to draw a straight line parallel to a given straight line.

Given: A point A and a line BC not containing it. To construct: A line through A parallel to BC. Construction: (1) Take a point D on BC. (2) Copy angle ADC into angle DAE with E and C on opposite sides of AD. The line EA is parallel to BC.