The Elements of Euclid for the Use of Schools and Colleges (Illustrated) E-Book

Isaac Todhunter

• Herausgeber: Balungi Francis
• Kategorie: Wissenschaft und neue Technologien
• Sprache: Englisch
• Veröffentlichungsjahr: 2021

Beschreibung

The subject of Plane Geometry is here presented to the student arranged in six books, and each book is subdivided into propositions. To assist the student in following the steps of the reasoning, references are given to the results already obtained which are required in the demonstration.

The books cover plane and solid Euclidean geometry, elementary number theory, and incommensurable lines. Elements is the oldest extant large-scale deductive treatment of mathematics. It has proven instrumental in the development of logic and modern science, and its logical rigor was not surpassed until the 19th century.

Euclid's Elements has been referred to as the most successful and influential textbook ever written. It was one of the very earliest mathematical works to be printed after the invention of the printing press and has been estimated to be second only to the Bible in the number of editions published since the first printing in 1482, with the number reaching well over one thousand. For centuries, when the quadrivium was included in the curriculum of all university students, knowledge of at least part of Euclid's Elements was required of all students. Not until the 20th century, by which time its content was universally taught through other school textbooks, did it cease to be considered something all educated people had read

This Original Edition Contains SIX Books:

• BOOK I.1
• BOOK II.90
• BOOK III.119
• BOOK IV.184
• BOOK V.215
• BOOK VI.274
• BOOK XI.335
• BOOK XII.369
• NOTES ON EUCLID'S ELEMENTS.377
• THE FIRST BOOK.380
• THE SECOND BOOK.401
• THE THIRD BOOK.408
• THE FOURTH BOOK.417
• THE FIFTH BOOK.419
• THE SIXTH BOOK.425
• THE ELEVENTH BOOK.434
• THE TWELFTH BOOK.437
• APPENDIX.438

Leseprobe

THE ELEMENTS OF

EUCLID

FOR THE USE OF SCHOOLS AND COLLEGES;

COMPRISING THE FIRST SIX BOOKS AND PORTIONS OF THE ELEVENTH AND TWELFTH BOOKS;

WITH COLOURED DIAGRAMS, SYMBOLS, NOTES, AN APPENDIX, AND EXERCISES

BY

I.TODHUNTER M.A.,F.R.S.

NEW EDITION. LONDON:

1872

https://www.bit.ly/findyogenius

Contents

INTRODUCTORY REMARKS.

BOOK I.

BOOK II.

BOOK III.

BOOK IV.

BOOK V.

BOOK VI.

BOOK XI.

BOOK XII.

NOTES ON EUCLID'S ELEMENTS.

THE FIRST BOOK.

THE SECOND BOOK.

THE THIRD BOOK.

THE FOURTH BOOK.

THE FIFTH BOOK.

THE SIXTH BOOK.

THE ELEVENTH BOOK.

THE TWELFTH BOOK.

APPENDIX.

INTRODUCTORY REMARKS.

The subject of Plane Geometry is here presented to the student arranged in six books, and each book is subdivided into propositions. The propositions are of two kinds, problems and theorems. In a problem something is required to be done; in a theorem some new principle is asserted to be true.

A proposition consists of various parts. We have first the general enunciation of the problem or theorem; as for example, To describe an equilateral triangle on a given finite straight line, or Any two angles of a triangle are together less than two right angles. After the general enunciation follows the discussion of the proposition.

First, the enunciation is repeated and applied to the particular figure which is to be considered; as for example, Let AB be the given straight line: it is required to describe an equilateral triangle on AB. The construction then usually follows, which states the necessary straight lines and circles which must be drawn in order to constitute the solution of the problem, or to furnish assistance in the demonstration of the theorem. Lastly, we have the demonstration itself, which shews that the problem has been solved, or that the theorem is true. Sometimes, however, no construction is required; and sometimes the construction and demonstration are combined.

The demonstration is a process of reasoning in which we draw inferences from results already obtained. These results consist partly of truths established in former propositions, or admitted as obvious in commencing the subject, and partly of truths which follow from the construction that has been made, or which are given in the supposition of the proposition itself. The word hypothesis is used in the same sense as supposition.

To assist the student in following the steps of the reasoning, references are given to the results already obtained which are required in the demonstration. Thus I. 5 indicates that we appeal to the result established in the fifth proposition of the First Book; Constr. is sometimes used as an abbreviation of Construction, and Hyp. as an abbreviation of Hypothesis.

It is usual to place the letters q.e.f. at the end of the discussion of a problem, and the letters q.e.d. at the end of the discussion of a theorem, q.e.f. is an abbreviation for quod erat faciendum, that is, which was to be done; and q.e.d. is an abbreviation for quod erat demonstrandum, that is, which was to be proved.

EUCLID'S ELEMENTS.

BOOK I.

DEFINITIONS.

1. A point is that which has no parts, or which has no magnitude.

2. A line is length without breadth.

3. The extremities of a line are points.

4. A straight line is that which lies evenly between its extreme points.

5. A superficies is that which has only length and breadth.

6. The extremities of a superficies are lines.

7. A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies.

8. A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction.

​9. A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line.

Note. When several angles are at one point B, any one of them is expressed by three letters, of which the letter which is at the vertex of the angle, that is, at the point at which the straight lines that contain the angle meet one another, is put between the other two letters, and one of these two letters is somewhere on one of those straight lines, and the other letter on the other straight line. Thus, the angle which is contained by the

straight lines AB, CB is named the angle ABC, or CBA; the angle which is contained by the straight lines AB, DB is named the angle ABD, or DBA; and the angle which is contained by the straight lines DB, CB is named the angle DBC, or CBD; but if there be only one angle at a point, it may be expressed by a letter placed at that point; as the angle at E.

10. When a straight line standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.

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11. An obtuse angle is that which is greater than a right angle.

12. An acute angle is that which is less than a right angle.

​13. A term or boundary is the extremity of any thing.

14. A figure is that which is enclosed by one or more boundaries.

15. A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one

another:

16. And this point is called the centre of the circle.

17. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.

[A radius of a circle is a straight line drawn from the centre to the circumference.]

18. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter.

19. A segment of a circle is the figure contained by a straight line and the circumference which it cuts off.

20. Rectilineal figures are those which are contained by straight lines:

21. Trilateral figures, or triangles, by three straight lines:

22. Quadrilateral figures by four straight lines:

23. Multilateral figures, or polygons, by more than four straight lines.

24. Of three-sided figures,

An equilateral triangle is that which has three equal sides:

25. An isosceles triangle is that which has two sides equal:

26. A scalene triangle is that which has three unequal sides:

27. A right-angled triangle is that which has a right angle:

[The side opposite to the right angle in a right-angled

triangle is frequently called the hypotenuse.]

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28. An obtuse-angled triangle is that which has an obtuse angle:

29. An acute-angled triangle is that which has three acute angles.

Of four-sided figures,

30. A square is that which has all its sides equal, and all its angles right angles:

31. An oblong is that which has all its angles right angles, but not all its sides equal:

32. A rhombus is that which has all its sides equal, but its angles are not right angles:

33. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles:

34. All other four-sided figures besides these are called trapeziums.

35. Parallel straight lines are such as are in the same plane, and which being produced ever

so far both ways do not meet.

[Note. The terms oblong and rhomboid are not often used. Practically the following definitions are used. Any four-sided figure is called a quadrilateral. A line joining two opposite angles of a quadrilateral is called a diagonal. A quadrilateral which has its opposite sides parallel is called a parallelogram'. The words square and rhombus are used in the sense defined by Euclid; and the word rectangle is used instead of the word oblong.

Some writers propose to restrict the word trapezium to a quadrilateral which has two of its sides parallel; and it would certainly be convenient if this restriction were universally adopted.]

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POSTULATES.

Let it be granted,

1. That a straight line may be drawn from any one point to any other point:

2. That a terminated straight line may be produced to any length in a straight line:

3. And that a circle may be described from any centre, at any distance from that centre.

AXIOMS.

1. Things which are equal to the same thing are equal to one another.

2. If equals be added to equals the wholes are equal.

3. If equals be taken from equals the remainders are equal.

4. If equals be added to unequals the wholes are unequal.

5. If equals be taken from unequals the remainders are unequal.

6. Things which are double of the same thing are equal to one another.

7. Things which are halves of the same thing are equal to one another.

8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

9. The whole is greater than its part.

10. Two straight lines cannot enclose a space.

11. All right angles are equal to one another.

12. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles.

PROPOSITION 1. PROBLEM.

To describe an equilateral triangle on a given finite straight line.

Let AB be the given straight line; it is required to describe an equilateral triangle on AB.

From the centre A at the distance AB describe the circle BCD. [Postulate 3.

From the centre B, at the distance BA, describe the circle ACE. [Postulate 3.

From the point C, at which the circles cut one another, draw the straight lines CA and CB to the points A and B. [Post. 1.

ABC shall be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal to AB. [Definition 15.

And because the point B is the centre of the circle ACE, BC is equal to BA. [Definition 15.

But it has been shewn that CA is equal to AB; therefore CA and CB are each of them equal to AB.

But things which are equal to the same thing are equal to one another. [Axiom 1.Therefore CA is equal to CB.Therefore CA, AB, BC are equal to one another.

Wherefore the triangle ABC is equilateral, [Def. 24. and it is described on the given straight line AB. Q.E.F.

PROPOSITION 2. PROBLEM.

From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.

From the point A to B draw the straight line AB; [Post.1.

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and on it describe the equilateral triangle DAB, [I. 1.

and produce the straight lines DA, DB to E and F. [Post. 2.

From the centre B, at the distance BC, describe the circle CGH, meeting DF at G. [Post.3.

From the centre D, at the distance DG, describe the circle GKL, meeting DE at L.[Post.3.AL shall be equal to BC.

Because the point B is the centre of the circle CGH, BC is equal to BG. [Definition 15.And because the point D is the centre of the circle GKL, DL is equal to DG; [Definition 15.

and DA, DB parts of them are equal; [Definition 24.

therefore the remainder AL is equal to the remainder BG.[Axiom3. But it has been shewn that BC is equal to BG;

therefore AL and BC are each of them equal to BG. But things which are equal to the same thing are equal to one another. [Axiom1.Therefore AL is equal to BC.

Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. q.e.f.

PROPOSITION 3. PROBLEM.

From the greater of two given straight lines to cut off a part equal to the less.

Let AB and C be the two given straight lines, of which ​AB

is the greater: it is required to cut off from AB the greater, a part equal to C the less.

From the point A draw the straight line AD equal to C; [I. 2. and from the centre A, at the distance AD, describe the circle DEF meeting AB at E. [Postulate 3.AE shall be equal to C. Because the point A is the centre of the circle DEF, AE is equal to AD. [Definition 15.But C is equal to AD. [Construction.Therefore AE and C are each of them equal to AD.Therefore AE is equal to C. [Axiom 1.

Wherefore from AB the greater of two given straight lines a part AE has been cut off equal to C the less. q.e.f.

PROPOSITION 4. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides

are opposite.

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Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, and the angle BAC equal to the angle EDF: the base BC shall be equal to the base EF, and the triangle ABC to the triangle DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

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For if the triangleABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB on the straight line DE, the point B will coincide with the point E, because AB is equal to DE, [Hyp.

And, AB coinciding with DE, AC will fall on DF, because the angle BAC is equal to the angle EDF. [Hypothesis.

Therefore also the point C will coincide with the point F, because AC is equal to DF. [Hypothesis.

But the point B was shewn to coincide with the point E, therefore the base BC will coincide with the base EF;

because, B coinciding with E and C with F, if the base BC does not coincide with the base EF, two straight lines will enclose a space; which is impossible. [Axiom 10.

Therefore the base BC coincides with the base EF, and is equal to it. [Axiom 8.

Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it. [Axiom 8.

And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

Wherefore, if two triangles &c. q.e.d.

PROPOSITION 5. THEOREM.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.

Let ABC be an isosceles triangle, having the side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E: the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD take any point F,and from AE the greater cut off AG equal to AF the less, [I.3.and join FC, GB.

Because AF is equal to AG [Constr.and AB to AC, [Hypothesis.the two sides FA, AC are equal to the two sides GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the base GB, and the triangle AFC to the triangle AGB, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, namely the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4. And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, [Hypothesis. the remainder BF is equal to the remainder CG. [Axiom 3. And FC was shewn to be equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB, each to each; and the angle BFC was shewn to be equal to the angle CGB; therefore the triangles BFC, CGB are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle FBC to the angle GCB, and the angle BCF to the angle CBG. [I. 4. And since it has been shewn that the whole angle ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF are also equal; therefore the remaining angle ABC is equal to the remaining angle ACB, which are the angles at the base of the triangle ABC. [Axiom 3. And it has also been shewn that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base. Wherefore, the angles &c. q.e.d.

Corollary. Hence every equilateral triangle is also equiangular.

PROPOSITION 6. THEOREM.

If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal

angles, shall be equal to one another.

Let ABC be a triangle, having the angle ABC equal to the angle ACB: the side AC shall be equal to the side AB. For if AC be not equal to AB, one of them must be greater than the other.

Let AB be the greater, and from it cut off DB equal to AC the less, [I. 3. and join DC. Then, because in the triangles DBC, ACB, DB is equal to AC, [Construction. and BC is common to both, the two sides DB, BC are equal to the two sides AC, CB each to each; and the angle DBC is equal to the angle ACB; [Hypothesis. therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB, [I. 4. the less to the greater; which is absurd. [Axiom 9.

Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles &c. q.e.d.

Corollary. Hence every equiangular triangle is also equilateral.

PROPOSITION 7. THEOREM.

On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the

other extremity equal to one another.

If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB,

having their sides CA, DA, which are terminated at the extremity A of the base, equal to one another, and likewise their sides CB, DB, which are terminated at B equal to one another.

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Join CD. In the case in which the vertex of each

triangle is without the other triangle;

because AC is equal to AD,[Hypothesis. the angle ACD is equal to the angle ADC.[I. 5. But the angle ACD is greater than the angle BCD,[Ax. 9. therefore the angle ADC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD.

Again, because BC is equal to BD,[Hypothesis. the angle BDC is equal to the angle BCD.[I. 5. But it has been shewn to be greater; which is impossible. But if one of the vertices as D, be within the other triangle ACB, produce AC, AD to E, F. Then because AC is equal to AD, in the triangle ACD, [Hyp. the angles ECD, FDC, on the other side of the base CD, are equal to one another. [I. 5. But the angle ECD is greater than the angle BCD, [Axiom 9. therefore the angle FDC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD.

Again, because BC is equal to BD, [Hypothesis. the angle BDC is equal to the angle BCD. [I. 5. But it has been shewn to be greater; which is impossible.

The case in which the vertex of one triangle is on a side of the other needs no demonstration.

Wherefore, on the same base &c. q.e.d.

PROPOSITION 8. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their ​bases equal, the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides, equal to them, of the other.

Let ABC, DEF be two triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each, namely AB to DE, and AC to DF, and also the base BC

equal to the base EF: the angle BAC shall be equal to the angle EDF.

For if the triangle ABC be applied to the triangle DEF, so that the point B may be on the point E, and the straight line BC on the straight line EF, the point C will also coincide with the point F, because BC is equal to EF. [Hyp.

Therefore, BC coinciding with EF, BA and AC will coincide with ED and DF.

For if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG; then on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity.  But this is impossible. [I. 7.  Therefore since the base BC coincides with the base EF, the sides BA, AC must coincide with the sides ED, DF. Therefore also the angle BAC coincides with the angle EDF, and is equal to it. [Axiom 8.

Wherefore, if two triangles &c. q.e.d.  

PROPOSITION 9. PROBLEM.

To bisect a given rectilineal angle, that is to divide it into

two equal angles.

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Let BAC be the given rectilineal angle: it is required to bisect it.

Take any point D in AB, and from AC cut off AE equal to AD; [I. 3. join DE, and on DE, on the side remote from A, describe the equilateral triangle DEF. [I. 1. Join AF. The straight line AF shall bisect the angle BAC. Because AD is equal to AE, [Construction. and AF is common to the two triangles DAF, EAF, the two sides DA, AF are equal to the two sides EA, AF, each to each; and the base DF is equal to the base EF; [Definition 24. therefore the angle DAF is equal to the angle EAF. [I. 8.

Wherefore the given rectilineal angle BAC is bisected by the straight line AF. q.e.f.

PROPOSITION 10. PROBLEM.

To bisect a given finite straight line, that is to divide it into two equal parts.

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Let AB be the given straight line; it is required to

divide it into two equal parts.

Describe on it an equilateral triangle ABC, [I. 1. and bisect the angle ACB by the straight line CD, meeting AB at D. [I. 9. AB shall be cut into two equal parts at the point D. Because AC is equal to CB, [Definition 24. and CD is common to the two triangles ACD, BCD, the two sides AC, CD are equal to the two sides BC, CD, each to each; and the angle ACD is equal to the angle BCD; [Constr. therefore the base AD is equal to the base DB. [I. 4.

Wherefore the given straight line AB is divided into two equal parts at the point D. q.e.f.

PROPOSITION 11. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

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Let AB be the given straight line, and C the given point in it: it is required to draw from the point C a straight line at right angles to AB.

Take any point D in AC and make CE equal to CD [I. 3 On DE describe the equilateral triangle DFE, [I. 1 and join CF.

The straight line CF drawn from the given point C shall be at right angles to the given straight line AB. Because DC is equal to CE, [Construction. and CF is common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; [Definition 24. therefore the angle DCF is equal to the angle ECF; [I. 8. and they are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [Definition 10. therefore each of the angles DCF, ECF is a right angle.

Wherefore from the given point C in the given straight line AB, CF has been drawn at right angles to AB. q.e.f.

Corollary. By the help of this problem it may be shewn that two straight lines cannot have a common segment.

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If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them.

From the point B draw BE at right angles to AB. Then, because ABC is a straight line, [Hypothesis. the angle CBE is equal to the angle EBA. [Definition 10. Also, because ABD is a straight line, [Hypothesis. the angle DBE is equal to the angle EBA.

Therefore the angle DBE is equal to the angle CBE, [Ax. 1. the less to the greater; which is impossible. [Axiom 9.

Wherefore two straight lines cannot have a common segment.

PROPOSITION 12. PROBLEM.

To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced to any length both ways, and let C be the given point without it: it is required to draw from the point C a straight line perpendicular to AB.

Take any point D on the other side of AB, and from the centre C, at the distance CD, describe the circle EGF, meeting AB at F and G. [Postulate 3. Bisect FG at H, [I. 10. and join CH.

The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB. Join CF, CG. Because FH is equal to HG, [Construction. and HC is common to the two triangles FHC, GHC; the two sides FH,HC are equal to the two sides GH, HC, each to each; and the base CF is equal to the base CG; [Definition 15. therefore the angle CHF is equal to the angle CHG; [I. 8. and they are adjacent angles.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [Def. 10.

Wherefore a perpendicular CH has been drawn to the given straight line AB from the given point C without it.  q.e.f.

PROPOSITION 13. THEOREM.

The angles which one straight line makes with another straight line on one side of it, either are two right angles, or are together equal to two right angles.

Let the straight line AB make with the straight line CD, on one side of it, the angles CBA, ABD: these either are two right angles, or are together equal to two right angles.

For if the angle CBA is equal to the angle ABD, each of them is a right angle. [Definition 10. But if not, from the point B draw BE at right angles to CD; [I. 11. therefore the angles CBE, EBD are two right angles, [Def. 10.

Now the angle CBE is equal to the two angles CBA, ABE; to each of these equals add the angle EBD; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [Axiom 2.

Again, the angle DBA is equal to the two angles DBE, EBA; to each of these equals add the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [Axiom 2. But the angles CBE, EBD have been shewn to be equal to the same three angles. Therefore the angles CBE, EBD are equal to the angles DBA, ABC [Axiom 1. But CBE, EBD are two right angles; therefore DBA, ABC are together equal to two right angles.

Wherefore, the angles &c. q.e.d.

PROPOSITION 14. THEOREM.

If, at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together

equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles: BD shall be in the same straight line with CB.

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For if BD be not in a the same straight line with CB, let BE be in the same straight line with it. Then because the straight line AB makes with the straight line CBE, on one side of it, the angles ABC, ABE, these angles are together equal to two right angles. [I. 13. But the angles ABC, ABD are also together equal to two right angles. [Hypothesis. Therefore the angles ABC, ABE are equal to the angles ABC, ABD.

From each of these equals take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, [Axiom 3. the less to the greater; which is impossible. Therefore BE is not in the same straight line with CB. And in the same manner it may be shewn that no other can be in the same straight line with it but BD; therefore BD is in the same straight line with CB.

Wherefore, if at a point &c. q.e.d.

PROPOSITION 15. THEOREM.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines AB, CD cut one another at the point E; the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED. Because the straight line AE makes with the straight line CD the angles CEA, AED, those angles are together equal to two right angles. [I. 13. Again, because the straight line DE makes with the straight line AB the angles AED, DEB, these also are together equal to two right angles. [I. 13. But the angles CEA, AED have been shewn to be together equal to two right angles. Therefore the angles CEA,AED are equal to the angles AED, DEB.

From each of these equals take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB. [Axiom 3.

In the same manner it may be shewn that the angle CEB is equal to the angle AED.

Wherefore, if two straight lines &c. q.e.d.

Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Corollary 2. And consequently, that all the angles made by any number of straight lines meeting at one point, are together equal to four right angles.

PROPOSITION 16. THEOREM.

If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.

Let ABC be a triangle, and let one side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC. Bisect AC at E, [I. 10. join BE and produce it to F, making EF equal to EB, [I. 3. and join FC. Because AE is equal to EC, and BE to EF; [Constr. the two sides AE, EB are equal to the two sides CE, EF, each to each; and the angle AEB is equal to the angle CEF, because they are opposite vertical angles; [I. 15. therefore the triangle AEB is equal to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; [I. 4. therefore the angle BAE is equal to the angle ECF.

But the angle ECD is greater than the angle ECF. [Axiom 9.

Therefore the angle ACD is greater than the angle BAE.

In the same manner if BC be bisected, and the side AC be produced to G, it may be shewn that the angle BCG, that is the angle ACD, is greater than the angle ABC. [I. 15.

Wherefore, if one side &c. q.e.d.

PROPOSITION 17. THEOREM.

Any two angles of a triangle are together less than two right angles.

Let ABC be a triangle: any two of its angles are

together less than two right angles.

Produce BC to D.Then because ACD is the exterior angle of the triangle ABC, it is greater than the interior opposite angle ABC. [I. 16. To each of these add the angle ACB

Therefore the angles ACD, ACB are greater than the angles ABC ACB. But the angles ACD, ACB are together equal to two right angles. [I. 13. Therefore the angles ABC, ACB are together less than two right angles.

In the same manner it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles.

Wherefore, any two angles &c. q.e.d.

PROPOSITION 18. THEOREM.

The greater side of every triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side AC is greater than the side AB: the angle ABC is also greater than

the angle ACB.

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Because AC is greater than AB, make AD equal to AB, [I. 3. and join BD.

Then, because ADB is the exterior angle of the triangle BDC, it is greater than the interior opposite angle DCB. [I. 16. But the angle ADB is equal to the angle ABD, [I. 5. because the side AD is equal to the side AB. [Constr. Therefore the angle ABD is also greater than the angle ACB. Much more then is the angle ABC greater than the angle ACB. [Axiom 9.

Wherefore, the greater side &c. q.e.d.

PROPOSITION 19. THEOREM.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

Let ABC be a triangle, of which the angle ABC is greater than the angle ACB: the side AC is also

greater than the side AB.

For if not, AC must be either equal to AB or less than AB. But AC is not equal to AB, for then the angle ABC would be equal to the angle ACB; [I. 5. but it is not; [Hypothesis. therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would be less than the angle ACB; [I. 18. but it is not; [Hypothesis. therefore AC is not less than AB.

And it has been shewn that AC is not equal to AB. Therefore AC is greater than AB.

Wherefore, the greater angle &c. q.e.d.

PROPOSITION 20. THEOREM.

Any two sides of a triangle are together greater than the

third side.

Let ABC be a triangle: any two sides of it are together greater than the third side; namely, BA, AC greater than BC; and AB, BC greater than AC; and BC, CA greater than AB.

Produce BA to D, making AD equal to AC, [I. 3. and join DC. Then, because AD is equal to AC, [Construction. the angle ADC is equal to the angle ACD. [I. 5. But the angle BCD is greater than the angle ACD. [Ax. 9. Therefore the angle BCD is greater than the angle BDC. And because the angle BCD of the triangle BCD is greater than its angle BDC, and that the greater angle is subtended by the greater side; [I. 19. therefore the side BD is greater than the side BC. But BD is equal to BA and AC. Therefore BA, AC are greater than BC.

In the same manner it may be shewn that AB, BC are greater than AC, and BC, CA greater than AB.

Wherefore, any two sides &c. q.e.d.

PROPOSITION 21. THEOREM.

If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD, DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC. Produce BD to meet AC at E. Because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than the side BE. [I. 20. To each of these add EC. Therefore BA, AC are greater than BE, EC.

Again; the two sides CE, ED of the triangle CED are greater than the third side CD. [I. 20. To each of these add DB. Therefore CE, EB are greater than CD, DB. But it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of any triangle is greater than the interior opposite angle, the exterior angle BDC of the triangle CDE is greater than the angle CED. [I. 16. For the same reason, the exterior angle CEB of the triangle ABE is greater than the angle BAE. But it has been shewn that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC.

Wherefore, if from the ends &c. q.e.d.

PROPOSITION 22. PROBLEM.

To make a triangle of which the sides shall be equal to

three given straight lines, but any two whatever of these must be greater than the third.

Let A, B, C be the three given straight lines, of which any two whatever are greater than the third; namely, A and B greater than C; A and C greater than B; and B and C greater than A: it is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

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Take a straight lineDE terminated at the point D, but unlimited towards E, and make DF equal to A, FG equal to B, and GH equal to C. [I. 3. From the centre F, at the distance FD, describe the circle DKL. [Post. 3. From the centre G, at the distance GH, describe the circle HLK, cutting the former circle at K. Join KF, KG. The triangle KFG shall have its sides equal to the three straight lines A,B, C. Because the point F is the centre of the circle DKL, FD is equal to FK. [Definition 15. But FD is equal to A. [Construction. Therefore FK is equal to A. [Axiom 1.

Again, because the point G is the centre of the circle HLK, GH is equal to GK [Definition 15. But GH is equal to C. [Construction. Therefore GK is equal to C. [Axiom 1. And FG is equal to B. [Construction. Therefore the three straight lines KF, FG, GK are equal to the three A, B, C.

Wherefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. q.e.f.

PROPOSITION 23. PROBLEM.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle: it is required to make at the given point A, in the given straight line AB, an angle equal to the given rectilineal angle DCE.In CD, CE take any points D, E, and join DE. Make the triangle AFG the sides of which shall be equal to the three straight lines CD, DE, EC; so that AF shall be equal to CD, AG to CE, and FG to DE. [I. 22. The angle FAG shall be equal to the angle DCE. Because FA, AG are equal to DC, CE, each to each, and the base FG equal to the base DE; [Construction. therefore the angle FAG is equal to the angle DCE. [I. 8.

Wherefore at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. q.e.f.

PROPOSITION 24. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other, the base of that which has the greater angle shall be greater than the base of the other.

Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, but the angle BAC greater than the angle EDF: the base BC shall be ​greater than the base EF.

Of the two sides DE, DF, let DE be the side which is

not greater than the other. At the point D in the straight line DE, make the angle EDG equal to the angle BAC, [I. 23. and make DG equal to AC or DF, [I. 3. and join EG, GF.

Because AB is equal to DE, [Hypothesis. and AC to DG; [Construction. the two sides BA, AC are equal to the two sides ED, DG, each to each; and the angle BAC is equal to the angle EDG; [Constr. therefore the base BC is equal to the base EG. [I. 4. And because DG is equal to DF, [Construction. the angle DGF is equal to the angle DFG. [I. 5. But the angle DGF is greater than the angle EGF. [Ax. 9.

Therefore the angle DFG is greater than the angle EGF. Much more then is the angle EFG greater than the angle EGF. [Axiom 9. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side, [I. 19. therefore the side EG is greater than the side EF. But EG was shewn to be equal to BC; therefore BC is greater than EF. Wherefore, if two triangles &c. q.e.d.

PROPOSITION 25. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one ​greater than the base of the other, the angle contained by the sides

of that which has the greater base, shall be greater than the angle contained by the sides equal to them, of the other.

Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, but the base BC greater than the base EF: the angle BAC shall be greater than the angle EDF.For if not, the angle BAC must be either equal to the angle EDF or less than the angle EDF. But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF; [I. 4. but it is not; [Hypothesis. therefore the angle BAC is not equal to the angle EDF.

Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF; [I. 24. but it is not; [Hypothesis. therefore the angle BAC is not less than the angle EDF. And it has been shewn that the angle BAC is not equal to the angle EDF. Therefore the angle BAC is greater than the angle EDF.

Wherefore, if two triangles &c. q.e.d.

PROPOSITION 26. THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other.

Let ABC, DEF be two triangles, which have the angles ABC, BCA equal to the angles DEF, EFD, each ​to