The First Six Books of the Elements of Euclid

INTRODUCTION.

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i. A point is that which has position but not dimensions.

A geometrical magnitude which has three dimensions, that is, length, breadth, and thickness, is a solid; that which has two dimensions, such as length and breadth, is a surface; and that which has but one dimension is a line.But a point is neither a solid, nor a surface, nor a line; hence it has no dimensions—that is, it has neither length, breadth, nor thickness.

ii. A line is length without breadth.

A line is space of one dimension.If it had any breadth, no matter how small, it would be space of two dimensions; and if in addition it had any thickness it would be space of three dimensions; hence a line has neither breadth nor thickness.

iii. The intersections of lines and their extremities are points.

iv. A line which lies evenly between its extreme points is called a straight or right line, such as AB.

If a point move without changing its direction it will describe a right line.The direction in which a point moves in called its “sense.”If the moving point continually changes its direction it will describe a curve; hence it follows that only one right line can be drawn between two points.The following Illustration is due to Professor Henrici:—“If we suspend a weight by a string, the string becomes stretched, and we say it is straight, by which we mean to express that it has assumed a peculiar definite shape.If we mentally abstract from this string all thickness, we obtain the notion of the simplest of all lines, which we call a straight line.”

v. A surface is that which has length and breadth.

A surface is space of two dimensions.It has no thickness, for if it had any, however small, it would be space of three dimensions.

vi. When a surface is such that the right line joining any two arbitrary points in it lies wholly in the surface, it is called a plane

A plane is perfectly flat and even, like the surface of still water, or of a smooth floor.—Newcomb

vii. Any combination of points, of lines, or of points and lines in a plane, is called a plane figure. If a figure be formed of points only it is called a stigmatic figure; and if of right lines only, a rectilineal figure.

viii. Points which lie on the same right line are called collinear points. A figure formed of collinear points is called a row of points.

ix. The inclination of two right lines extending out from one point in different directions is called a rectilineal angle.

x. The two lines are called the legs, and the point the vertex of the angle.

A right line drawn from the vertex and turning about it in the plane of the angle, from the position of coincidence with one leg to that of coincidence with the other, is said to turn through the angle, and the angle is the greater as the quantity of turning is the greater.Again, since the line may turn from one position to the other in either of two ways, two angles are formed by two lines drawn from a point.

Thus if AB, AC be the legs, a line may turn from the position AB to the position AC in the two ways indicated by the arrows.The smaller of the angles thus formed is to be understood as the angle contained by the lines. The larger, called a re-entrant angle, seldom occurs in the “Elements.”

xi. Designation of Angles.—A particular angle in a figure is denoted by three letters, as BAC, of which the middle one, A, is at the vertex, and the other two along the legs. The angle is then read BAC.

xii. The angle formed by joining two or more angles together is called their sum. Thus the sum of the two angles ABC, PQR is the angle AB′R,

formed by applying the side QP to the side BC, so that the vertex Q shall fall on the vertex B, and the side QR on the opposite side of BC from BA

xiii. When the sum of two angles BAC, CAD is such that the legs BA, AD form one right line, they are called supplements of each other.

Hence, when one line stands on another, the two angles which it makes on the same side of that on which it stands are supplements of each other.

xivWhen one line stands on another, and makes the adjacent angles at both sides of itself equal, each of the angles is called a right angle, and the line which stands on the other is called a perpendicular to it.

Hence a right angle is equal to its supplement.

xv. An acute angle is one which is less than a right angle, as A.

xvi. An obtuse angle is one which is greater than a right angle, as BAC.

The supplement of an acute angle is obtuse, and conversely, the supplement of an obtuse angle is acute.

xvii. When the sum of two angles is a right angle, each is called the complement of the other. Thus, if the angle BAC be right, the angles BAD, DAC are complements of each other.

xviii. Three or more right lines passing through the same point are called concurrent lines.

xix. A system of more than three concurrent lines is called a pencil of lines. Each line of a pencil is called a ray, and the common point through which the rays pass is called the vertex.

xx. A triangle is a figure formed by three right lines joined end to end. The three lines are called its sides.

xxi. A triangle whose three sides are unequal is said to be scalene, as A; a triangle having two sides equal, to be isosceles, as B; and and having all its sides equal, to be equilateral, as C.

xxii. A right-angled triangle is one that has one of its angles a right angle, as D. The side which subtends the right angle is called the hypotenuse.

xxiii. An obtuse-angled triangle is one that has one of its angles obtuse, as E.

xxiv. An acute-angled triangle is one that has its three angles acute, as F.

xxv. An exterior angle of a triangle is one that is formed by any side and the continuation of another side.

Hence a triangle has six exterior angles; and also each exterior angle is the supplement of the adjacent interior angle.

xxvi. A rectilineal figure bounded by more than three right lines is usually called a polygon.

xxvii. A polygon is said to be convex when it has no re-entrant angle.

xxviii. A polygon of four sides is called a quadrilateral

xxix. A quadrilateral whose four sides are equal is called a lozenge

xxx. A lozenge which has a right angle is called a square

xxxi. A polygon which has five sides is called a pentagon; one which has six sides, a hexagon, and so on.

xxxii. A circle is a plane figure formed by a curved line called the circumference, and is such that all right lines drawn from a certain point within the figure to the circumference are equal to one another. This point is called the centre.

xxxiii. A radius of a circle is any right line drawn from the centre to the circumference, such as CD.

xxxiv. A diameter of a circle is a right line drawn through the centre and terminated both ways by the circumference, such as AB.

From the definition of a circle it follows at once that the path of a movable point in a plane which remains at a constant distance from a fixed point is a circle; also that any point P in the plane is inside, outside, or on the circumference of a circle according as its distance from the centre is less than, greater than, or equal to, the radius.

Let it be granted that—

i. A right line may be drawn from any one point to any other point.

When we consider a straight line contained between two fixed points which are its ends, such a portion is called a finite straight line

ii. A terminated right line may be produced to any length in a right line.

Every right line may extend without limit in either direction or in both.It is in these cases called an indefinite line.By this postulate a finite right line may be supposed to be produced, whenever we please, into an indefinite right line.

iii. A circle may be described from any centre, and with any distance from that centre as radius.

If there be two points A and B, and if with any instruments, such as a ruler and pen, we draw a line from A to B, this will evidently have some irregularities, and also some breadth and thickness.Hence it will not be a geometrical line no matter how nearly it may approach to one.This is the reason that Euclid postulates the drawing of a right line from one point to another.For if it could be accurately done there would be no need for his asking us to let it be granted.Similar observations apply to the other postulates.It is also worthy of remark that Euclid never takes for granted the doing of anything for which a geometrical construction, founded on other problems or on the foregoing postulates, can be given.

i. Things which are equal to the same, or to equals, are equal to each other.

Thus, if there be three things, and if the first, and the second, be each equal to the third, we infer by this axiom that the first is equal to the second.This axiom relates to all kinds of magnitude. The same is true of Axioms ii. , iii. , iv. , v. , vi. , vii. , ix. ; but viii. , x. , xi. , xii, are strictly geometrical.

ii. If equals be added to equals the sums will be equal.

iii. If equals be taken from equals the remainders will be equal.

iv. If equals be added to unequals the sums will be unequal.

v. If equals be taken from unequals the remainders will be unequal.

vi. The doubles of equal magnitudes are equal.

vii. The halves of equal magnitudes are equal.

viii. Magnitudes that can be made to coincide are equal.

The placing of one geometrical magnitude on another, such as a line on a line, a triangle on a triangle, or a circle on a circle, &c. , is called superpositionThe superposition employed in Geometry is only mental, that is, we conceive one magnitude placed on the other; and then, if we can prove that they coincide, we infer, by the present axiom, that they are equal.Superposition involves the following principle, of which, without explicitly stating it, Euclid makes frequent use:—“Any figure may be transferred from one position to another without change of form or size.”

ix. The whole is greater than its part.

This axiom is included in the following, which is a fuller statement:—

ix′. The whole is equal to the sum of all its parts.

x. Two right lines cannot enclose a space.

This is equivalent to the statement, “If two right lines have two points common to both, they coincide in direction,” that is, they form but one line, and this holds true even when one of the points is at infinity.

xi. All right angles are equal to one another.

This can be proved as follows:—Let there be two right lines AB, CD, and two perpendiculars to them, namely, EF, GH, then if AB, CD be made to coincide by superposition, so that the point E will coincide with G; then since a right angle is equal to its supplement, the line EF must coincide with GH. Hence the angle AEF is equal to CGH

xiiIf two right lines (AB, CD) meet a third line (AC), so as to make the sum of the two interior angles (BAC, ACD) on the same side less than two right angles, these lines being produced shall meet at some finite distance.

This axiom is the converse of Prop. xvii. , Book I.

Axioms. —“Elements of human reason,” according to Dugald Stewart, are certain general propositions, the truths of which are self-evident, and which are so fundamental, that they cannot be inferred from any propositions which are more elementary; in other words, they are incapable of demonstration. “That two sides of a triangle are greater than the third” is, perhaps, self-evident; but it is not an axiom, inasmuch as it can be inferred by demonstration from other propositions; but we can give no proof of the proposition that “things which are equal to the same are equal to one another,” and, being self-evident, it is an axiom.

Propositions which are not axioms are properties of figures obtained by processes of reasoning. They are divided into theorems and problems.

A Theorem is the formal statement of a property that may be demonstrated from known propositions. These propositions may themselves be theorems or axioms. A theorem consists of two parts, the hypothesis, or that which is assumed, and the conclusion, or that which is asserted to follow therefrom. Thus, in the typical theorem,

the hypothesis is that X is Y , and the conclusion is that Z is W.

Converse Theorems—Two theorems are said to be converse, each of the other, when the hypothesis of either is the conclusion of the other.Thus the converse of the theorem (i.) is—

From the two theorems (i) and (ii) we may infer two others, called their contrapositives. Thus the contrapositive

The theorem (iv.) is called the obverse of (i.), and (iii.)the obverse of (ii.) .

A Problem is a proposition in which something is proposed to be done, such as a line to be drawn, or a figure to be constructed, under some given conditions.

The Solution of a problem is the method of construction which accomplishes the required end.

The Demonstration is the proof, in the case of a theorem, that the conclusion follows from the hypothesis; and in the case of a problem, that the construction accomplishes the object proposed.

The Enunciation of a problem consists of two parts, namely, the data, or things supposed to be given, and the quaesita, or things required to be done.

Postulates are the elements of geometrical construction, and occupy the same relation with respect to problems as axioms do to theorems.

A Corollary is an inference or deduction from a proposition.

A Lemma is an auxiliary proposition required in the demonstration of a principal proposition.

A Secant or Transversal is a line which cuts a system of lines, a circle, or any other geometrical figure.

Congruent figures are those that can be made to coincide by superposition. They agree in shape and size, but differ in position. Hence it follows, by Axiom viii. , that corresponding parts or portions of congruent figures are congruent, and that congruent figures are equal in every respect.

Rule of Identity. —Under this name the following principle will be sometimes referred to:—“If there is but one X and one Y , then, from the fact that X is Y , it necessarily follows that Y is X.”—Syllabus.

Sol.—With A as centre, and AB as radius, describe the circle BCD (Post. iii). With B as centre, and BA as radius, describe the circle ACE, cutting the former circle in C. Join CA, CB (Post. i). Then ABC is the equilateral triangle required.

Dem.—Because A is the centre of the circle BCD, AC is equal to AB (Def. xxxii). Again, because B is the centre of the circle ACE, BC is equal to BA. Hence we have proved.

But things which are equal to the same are equal to one another (Axiom i); therefore AC is equal to BC; therefore the three lines AB, BC, CA are equal to one another. Hence the triangle ABC is equilateral (Def. xxi); and it is described on the given line AB, which was required to be done.

1. What is the datum in this proposition?

2. What is the quaesitum?

3. What is a finite right line?

4. What is the opposite of finite?

5. In what part of the construction is the third postulate quoted? and for what purpose? Where is the first postulate quoted?

6. Where is the first axiom quoted?

7. What use is made of the definition of a circle? What is a circle?

8. What is an equilateral triangle?

The following exercises are to be solved when the pupil has mastered the First Book:—

1. If the lines AF, BF be joined, the figure ACBF is a lozenge.

2. If AB be produced to D and E, the triangles CDF and CEF are equilateral.

3. If CA, CB be produced to meet the circles again in G and H, the points G, F, H are collinear, and the triangle GCH is equilateral.

4. If CF be joined, CF² = 3AB²

5. Describe a circle in the space ACB, bounded by the line AB and the two circles.

Sol.—Join AB (Post. i); on AB describe the equilateral triangle ABD [i]. With B as centre, and BC as radius, describe the circle ECH (Post iii). Produce DB to meet the circle ECH in E (Post. ii). With D as centre, and DE as radius, describe the circle EFG (Post. iii). Produce DA to meet this circle in F. AF is equal to BC.

Dem.—Because D is the centre of the circle EFG, DF is equal to DE (Def. xxxii). And because DAB is an equilateral triangle, DA is equal to DB (Def. xxi). Hence we have

and taking the latter from the former, the remainder AF is equal to the remainder BE (Axiom iii). Again, because B is the centre of the circle ECH, BC is equal to BE; and we have proved that AF is equal to BE; and things which are equal to the same thing are equal to one another (Axiom i). Hence AF is equal to BC. Therefore from the given point A the line AF has been drawn equal to BC

It is usual with commentators on Euclid to say that he allows the use of the rule and compass Were such the case this Proposition would have been unnecessary.The fact is, Euclid’s object was to teach Theoretical and not Practical Geometry, and the only things he postulates are the drawing of right lines and the describing of circles.If he allowed the mechanical use of the rule and compass he could give methods of solving many problems that go beyond the limits of the “geometry of the point, line, and circle.”—See Notes D, F at the end of this work.

1. Solve the problem when the point A is in the line BC itself.

2. Inflect from a given point A to a given line BC a line equal to a given line.State the number of solutions.

Sol.—From A, one of the extremities of AB, draw the right line AD equal to C [ii]; and with A as centre, and AD as radius, describe the circle EDF (Post. iii) cutting AB in E. AE shall be equal to C.

Dem.—Because A is the centre of the circle EDF, AE is equal to AD (Def. xxxii), and C is equal to AD (const.) ; and things which are equal to the same are equal to one another (Axiom i); therefore AE is equal to C. Wherefore from AB, the greater of the two given lines, a part, AE, has been out off equal to C, the less.

1. What previous problem is employed in the solution of this?

2. What postulate?

3. What axiom in the demonstration?

4. Show how to produce the less of two given lines until the whole produced line becomes equal to the greater.

If two triangles (BAC, EDF) have two sides (BA, AC) of one equal respectively to two sides (ED, DF) of the other, and have also the angles (A, D) included by those sides equal, the triangles shall be equal in every respect—that is, their bases or third sides (BC, EF) shall be equal, and the angles (B, C) at the base of one shall be respectively equal to the angles (E, F) at the base of the other; namely, those shall be equal to which the equal sides are opposite.

Dem.—Let us conceive the triangle BAC to be applied to EDF, so that the point A shall coincide with D, and the line AB with DE, and that the point C shall be on the same side of DE as F; then because AB is equal to DE, the point B shall coincide with E. Again, because the angle BAC is equal to the angle EDF, the line AC shall coincide with DF; and since AC is equal to DF (hyp.) , the point C shall coincide with F; and we have proved that the point B coincides with E. Hence two points of the line BC coincide with two points of the line EF; and since two right lines cannot enclose a space, BC must coincide with EF. Hence the triangles agree in every respect; therefore BC is equal to EF, the angle B is equal to the angle E, the angle C to the angle F, and the triangle BAC to the triangle EDF

1. How many parts in the hypothesis of this Proposition? Ans. Three. Name them.

2. How many in the conclusion? Name them.

3. What technical term is applied to figures which agree in everything but position? Ans. They are said to be congruent.

4. What is meant by superposition?

5. What axiom is made use of in superposition?

6. How many parts in a triangle? Ans. Six; namely, three sides and three angles.

7. When it is required to prove that two triangles are congruent, how many parts of one must be given equal to corresponding parts of the other? Ans. In general, any three except the three angles.This will be established in Props. viii and xxvi, taken along with iv

8. What property of two lines having two common points is quoted in this Proposition? They must coincide.

1. The line that bisects the vertical angle of an isosceles triangle bisects the base perpendicularly.

2. If two adjacent sides of a quadrilateral be equal, and the diagonal bisects the angle between them, their other sides are equal.

3. If two lines be at right angles, and if each bisect the other, then any point in either is equally distant from the extremities of the other.

4. If equilateral triangles be described on the sides of any triangle, the distances between the vertices of the original triangle and the opposite vertices of the equilateral triangles are equal.(This Proposition should be proved after the student has read Prop. xxxii)

The angles (ABC, ACB) at the base (BC) of an isosceles triangle are equal to one another, and if the equal sides (AB, AC) be produced, the external angles (DEC, ECB) below the base shall be equal.

Dem.—In BD take any point F, and from AE, the greater, cut off AG equal to AF [iii]. Join BG, CF (Post. i). Because AF is equal to AG (const.) , and AC is equal to AB (hyp.) , the two triangles FAC, GAB have the sides FA, AC in one respectively equal to the sides GA, AB in the other; and the included angle A is common to both triangles.Hence [iv] the base FC is equal to GB, the angle AFC is equal to AGB, and the angle ACF is equal to the angle ABG.

Again, because AF is equal to AG (const.) , and AB to AC (hyp.) , the remainder, BF, is equal to CG (Axiom iii); and we have proved that FC is equal to GB, and the angle BFC equal to the angle CGB. Hence the two triangles BFC, CGB have the two sides BF, FC in one equal to the two sides CG, GB in the other; and the angle BFC contained by the two sides of one equal to the angle CGB contained by the two sides of the other. Therefore [iv] these triangles have the angle FBC equal to the angle GCB, and these are the angles below the base. Also the angle FCB equal to GBC; but the whole angle FCA has been proved equal to the whole angle GBA. Hence the remaining angle ACB is equal to the remaining angle ABC, and these are the angles at the base.

Observation.—The great difficulty which beginners find in this Proposition is due to the fact that the two triangles ACF, ABG overlap each other.The teacher should make these triangles separate, as in the annexed diagram, and point out the corresponding parts thus:—

The student should also be shown how to apply one of the triangles to the other, so as to bring them into coincidence. Similar Illustrations may be given of the triangles BFC, CGB

The following is a very easy proof of this Proposition. Conceive the △ ACB to be turned, without alteration, round the line AC, until it falls on the other side. Let ACD be its new position; then the angle ADC of the displaced triangle is evidently equal to the angle ABC, with which it originally coincided. Again, the two △s BAC, CAD have the sides BA, AC of one respectively equal to the sides AC, AD of the other, and the included angles equal; therefore [iv] the angle ACB opposite to the side AB is equal to the angle ADC opposite to the side AC; but the angle ADC is equal to ABC; therefore ACB is equal to ABC

Cor.—Every equilateral triangle is equiangular.

Def—A line in any figure, such as AC in the preceding diagram, which is such that, by folding the plane of the figure round it, one part of the diagram will coincide with the other, is called an axis of symmetry of the figure.

1. Prove that the angles at the base are equal without producing the sides. Also by producing the sides through the vertex.

2. Prove that the line joining the point A to the intersection of the lines CF and BG is an axis of symmetry of the figure.

3. If two isosceles triangles be on the same base, and be either at the same or at opposite sides of it, the line joining their vertices is an axis of symmetry of the figure formed by them.

4. Show how to prove this Proposition by assuming as an axiom that every angle has a bisector.

5. Each diagonal of a lozenge is an axis of symmetry of the lozenge.

6. If three points be taken on the sides of an equilateral triangle, namely, one on each side, at equal distances from the angles, the lines joining them form a new equilateral triangle.

Dem.—If AB, AC are not equal, one must be greater than the other. Suppose AB is the greater, and that the part BD is equal to AC. Join CD (Post. i). Then the two triangles DBC, ACB have BD equal to AC, and BC common to both. Therefore the two sides DB, BC in one are equal to the two sides AC, CB in the other; and the angle DBC in one is equal to the angle ACB in the other (hyp).Therefore [iv] the triangle DBC is equal to the triangle ACB—the less to the greater, which is absurd; hence AC, AB are not unequal, that is, they are equal.

1. What is the hypothesis in this Proposition?

2. What Proposition is this the converse of?

3. What is the obverse of this Proposition?

4. What is the obverse of Prop. v?

5. What is meant by an indirect proof?

6. How does Euclid generally prove converse Propositions?

7. What false assumption is made in the demonstration?

8. What does this assumption lead to?

If two triangles (ACB, ADB) on the same base (AB) and on the same side of it have one pair of conterminous sides (AC, AD) equal to one another, the other pair of conterminous sides (BC, BD) must be unequal.

Dem.—1. Let the vertex of each triangle be without the other. Join CD. Then because AD is equal to AC (hyp.) , the triangle ACD is isosceles; therefore [v] the angle ACD is equal to the angle ADC; but ADC is greater than BDC (Axiom ix); therefore ACD is greater than BDC: much, more is BCD greater than BDC. Now if the side BD were equal to BC, the angle BCD would be equal to BDC [v]; but it has been proved to be greater. Hence BD is not equal to BC.

2. Let the vertex of one triangle ADB fall within the other triangle ACB. Produce the sides AC, AD to E and F. Then because AC is equal to AD (hyp.) , the triangle ACD is isosceles, and [v] the external angles ECD, FDC at the other side of the base CD are equal; but ECD is greater than BCD (Axiom ix). Therefore FDC is greater than BCD: much more is BDC greater than BCD; but if BC were equal to BD, the angle BDC would be equal to BCD [v]; therefore BC cannot be equal to BD.

3. If the vertex D of the second triangle fall on the line BC, it is evident that BC and BD are unequal.

1. What use is made of Prop. vii? Ans. As a lemma to Prop. viii

2. In the demonstration of Prop. vii the contrapositive of Prop. v occurs; show where.

3. Show that two circles can intersect each other only in one point on the same side of the line joining their centres, and hence that two circles cannot have more than two points of intersection.

If two triangles (ABC, DEF) have two sides (AB, AC) of one respectively equal to two sides (DE, DF) of the other, and have also the base (BC) of one equal to the base (EF) of the other; then the two triangles shall be equal, and the angles of one shall be respectively equal to the angles of the other—namely, those shall be equal to which the equal sides are opposite.

Dem.—Let the triangle ABC be applied to DEF, so that the point B will coincide with E, and the line BC with the line EF; then because BC is equal to EF, the point C shall coincide with F. Then if the vertex A fall on the same side of EF as the vertex D, the point A must coincide with D; for if not, let it take a different position G; then we have EG equal to BA, and BA is equal to ED (hyp.) . Hence (Axiom i) EG is equal to ED: in like manner, FG is equal to FD, and this is impossible [vii]. Hence the point A must coincide with D, and the triangle ABC agrees in every respect with the triangle DEF; and therefore the three angles of one are respectively equal to the three angles of the other—namely, A to D, B to E, and C to F, and the two triangles are equal.

This Proposition is the converse of iv, and is the second case of the congruence of triangles in the Elements.

Philo’s Proof.—Let the equal bases be applied as in the foregoing proof, but let the vertices be on the opposite sides; then let BGC be the position which EDF takes. Join AGThen because BG = BA, the angle BAG = BGA. In like manner the angle CAG = CGAHence the whole angle BAC = BGC; but BGC = EDF therefore BAC = EDF

Sol.—In AB take any point D, and cut off [iii] AE equal to AD. Join DE (Post. i), and upon it, on the side remote from A, describe the equilateral triangle DEF [i] Join AF. AF bisects the given angle BAC

Dem.—The triangles DAF, EAF have the side AD equal to AE (const.) and AF common; therefore the two sides DA, AF are respectively equal to EA, AF, and the base DF is equal to the base EF, because they are the sides of an equilateral triangle (Def. xxi).Therefore [viii] the angle DAF is equal to the angle EAF; hence the angle BAC is bisected by the line AF.

Cor.—The line AF is an axis of symmetry of the figure.

1. Why does Euclid describe the equilateral triangle on the side remote from A?

2. In what case would the construction fail, if the equilateral triangle were described on the other side of DE?

1. Prove this Proposition without using Prop. viii

2. Prove that AF is perpendicular to DE

3. Prove that any point in AF is equally distant from the points D and E

4. Prove that any point in AF is equally distant from the lines AB, AC

Sol.—Upon AB describe an equilateral triangle ACB [i].Bisect the angle ACB by the line CD [ix], meeting AB in D, then AB is bisected in D.

Dem.—The two triangles ACD, BCD, have the side AC equal to BC, being the sides of an equilateral triangle, and CD common. Therefore the two sides AC, CD in one are equal to the two sides BC, CD in the other; and the angle ACD is equal to the angle BCD (const.) . Therefore the base AD is equal to the base DB [iv]. Hence AB is bisected in D.

1. Show how to bisect a finite right line by describing two circles.

2. Every point equally distant from the points A, B is in the line CD

From a given point (C) in a given right line (AB) to draw a right line perpendicular to the given line.

Sol.—In AC take any point D, and make CE equal to CD [iii]. Upon DE describe an equilateral triangle DFE [i]. Join CF. Then CF shall be at right angles to AB

Dem.—The two triangles DCF, ECF have CD equal to CE (const.) and CF common; therefore the two sides CD, CF in one are respectively equal to the two sides CE, CF in the other, and the base DF is equal to the base EF, being the sides of an equilateral triangle (Def. xxi.); therefore [viii.] the angle DCE is equal to the angle ECF, and they are adjacent angles. Therefore (Def. xiii.) each of them is a right angle, and CF is perpendicular to AB at the point C.

1. The diagonals of a lozenge bisect each other perpendicularly.

2. Prove Prop. xi. without using Prop. viii

3. Erect a line at right angles to a given line at one of its extremities without producing the line.

4. Find a point in a given line that shall be equally distant from two given points.

5. Find a point in a given line such that, if it be joined to two given points on opposite sides of the line, the angle formed by the joining lines shall be bisected by the given line.

6. Find a point that shall be equidistant from three given points.

Sol.—Take any point D on the other side of AB, and describe (Post. iii.) a circle, with C as centre, and CD as radius, meeting AB in the points F and G. Bisect FG in H [x.] . Join CH (Post. i.) . CH shall be at right angles to AB

Dem.—Join CF, CG. Then the two triangles FHC, GHC have FH equal to GH (const.) , and HC common; and the base CF equal to the base CG, being radii of the circle FDG (Def. xxxii.) . Therefore the angle CHF is equal to the angle CHG [viii.] , and, being adjacent angles, they are right angles (Def. xiii.) . Therefore CH is perpendicular to AB.

1. Prove that the circle cannot meet AB in more than two points.

2. If one angle of a triangle be equal to the sum of the other two, the triangle can be divided into the sum of two isosceles triangles, and the base is equal to twice the line from its middle point to the opposite angle.

The adjacent angles (ABC, ABD) which one right line (AB) standing on another (CD) makes with it are either both right angles, or their sum is equal to two right angles.

Dem.—If AB is perpendicular to CD, as in fig. 1, the angles ABC, ABD are right angles. If not, draw BE perpendicular to CD [xi.] . Now the angle CBA is equal to the sum of the two angles CBE, EBA (Def. xi.) . Hence, adding the angle ABD, the sum of the angles CBA, ABD is equal to the sum of the three angles CBE, EBA, ABD. In like manner, the sum of the angles CBE, EBD is equal to the sum of the three angles CBE, EBA, ABD. And things which are equal to the same are equal to one another. Therefore the sum of the angles CBA, ABD is equal to the sum of the angles CBE, EBD; but CBE, EBD are right angles; therefore the sum of the angles CBA, ABD is two right angles.

Or thus: Denote the angle EBA by θ; then evidently

Cor. 1. —The sum of two supplemental angles is two right angles.

Cor. 2. —Two right lines cannot have a common segment.

Cor. 3. —The bisector of any angle bisects the corresponding re-entrant angle.

Cor. 4. —The bisectors of two supplemental angles are at right angles to each other.

Cor. 5. —The angle EBA is half the difference of the angles CBA, ABD.

If at a point (B) in a right line (BA) two other right lines (CB, BD) on opposite sides make the adjacent angles (CBA, ABD) together equal to two right angles, these two right lines form one continuous line.

Dem.—If BD be not the continuation of CB, let BE be its continuation. Now, since CBE is a right line, and BA stands on it, the sum of the angles CBA, ABE is two right angles (xiii.) ; and the sum of the angles CBA, ABD is two right angles (hyp.) ; therefore the sum of the angles CBA, ABE is equal to the sum of the angles CBA, ABD. Reject the angle CBA, which is common, and we have the angle ABE equal to the angle ABD—that is, a part equal to the whole—which is absurd. Hence BD must be in the same right line with CB

Dem.—Because the line AE stands on CD, the sum of the angles CEA, AED is two right angles [xiii.] ; and because the line CE stands on AB, the sum of the angles BEC, CEA is two right angles; therefore the sum of the angles CEA, AED is equal to the sum of the angles BEC, CEA. Reject the angle CEA, which is common, and we have the angle AED equal to BEC. In like manner, the angle CEA is equal to DEB.

The foregoing proof may be briefly given, by saying that opposite angles are equal because they have a common supplement.

1. What problem is required in Euclid’s proof of Prop. xiii.?

2. What theorem? Ans. No theorem, only the axioms.

3. If two lines intersect, how many pairs of supplemental angles do they make?

4. What relation does Prop. xiv bear to Prop. xiii.?

5. What three lines in Prop. xiv are concurrent?

6. What caution is required in the enunciation of Prop. xiv.?

7. State the converse of Prop. xvProve it.

8. What is the subject of Props. xiii. , xiv. , xv? Ans. Angles at a point.

PROP. XVI. —Theorem

If any side (BC) of a triangle (ABC) be produced, the exterior angle (ACD) is greater than either of the interior non-adjacent angles.

Dem.—Bisect AC in E [x.] . Join BE (Post. i.) . Produce it, and from the produced part cut off EF equal to BE [iii]. Join CF. Now because EC is equal to EA (const.) , and EF is equal to EB, the triangles CEF, AEB have the sides CE, EF in one equal to the sides AE, EB in the other; and the angle CEF equal to AEB [xv.]Therefore [iv.] the angle ECF is equal to EAB; but the angle ACD is greater than ECF; therefore the angle ACD is greater than EAB.

In like manner it may be shown, if the side AC be produced, that the exterior angle BCG is greater than the angle ABC; but BCG is equal to ACD [xv.] . Hence ACD is greater than ABC. Therefore ACD is greater than either of the interior non-adjacent angles A or B of the triangle ABC

Cor. 1.—The sum of the three interior angles of the triangle BCF is equal to the sum of the three interior angles of the triangle ABC.

Cor. 2.—The area of BCF is equal to the area of ABC.

Cor. 3.—The lines BA and CF, if produced, cannot meet at any finite distance. For, if they met at any finite point X, the triangle CAX would have an exterior angle BAC equal to the interior angle ACX.

Dem.—Produce BC to D; then the exterior angle ACD is greater than ABC [xvi]: to each add the angle ACB, and we have the sum of the angles ACD, ACB greater than the sum of the angles ABC, ACB; but the sum of the angles ACD, ACB is two right angles [xiii]. Therefore the sum of the angles ABC, ACB is less than two right angles.

In like manner we may show that the sum of the angles A, B, or of the angles A, C, is less than two right angles.

Cor. 1.—Every triangle must have at least two acute angles.

Cor. 2.—If two angles of a triangle be unequal, the lesser must be acute.

Prove Prop. xvii without producing a side.

If in any triangle (ABC) one side (AC) be greater than another (AB), the angle opposite to the greater side is grater than the angle opposite to the less.