Elementa

Euclid

Elementa

Euclid · Elem 1 · perseus-eng2 · More by Euclid

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Original / primary: Perseus Eng2

1.def.1.p1

A point is that which has no part.

1.def.2.p2

A line is breadthless length.

1.def.3.p3

The extremities of a line are points.

1.def.4.p4

A straight line is a line which lies evenly with the points on itself.

1.def.5.p5

A surface is that which has length and breadth only.

1.def.6.p6

The extremities of a surface are lines.

1.def.7.p7

A plane surface is a surface which lies evenly with the straight lines on itself.

1.def.8.p8

A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.

1.def.9.p9

And when the lines containing the angle are straight, the angle is called rectilineal.

1.def.10.p10

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.

1.def.11.p11

An obtuse angle is an angle greater than a right angle.

1.def.12.p12

An acute angle is an angle less than a right angle.

1.def.13.p13

A boundary is that which is an extremity of anything.

1.def.14.p14

A figure is that which is contained by any boundary or boundaries.

1.def.15.p15

A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another;

1.def.16.p16

And the point is called the centre of the circle.

1.def.17.p17

A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle.

1.def.18.p18

A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.

1.def.19.p19

Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.

1.def.20.p20

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.

1.def.21.p21

Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute.

1.def.22.p22

Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.

1.def.23.p23

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.

1.post.1.p24

Let the following be postulated:

1.post.1.p25

To draw a straight line from any point to any point.

1.post.2.p26

To produce a finite straight line continuously in a straight line.

1.post.3.p27

To describe a circle with any centre and distance.

1.post.4.p28

That all right angles are equal to one another.

1.post.5.p29

That, if a straight line falling on two straight lines make 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.

1.comm.not.1.p30

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

1.comm.not.2.p31

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

1.comm.not.3.p32

If equals be subtracted from equals, the remainders are equal.

1.comm.not.7.p33

[7] Things which coincide with one another are equal to one another.

1.comm.not.8.p34

[8] The whole is greater than the part.

1.prop.1.p35

Enunciation On a given finite straight line to construct an equilateral triangle.

1.prop.1.p36

Proof. Let AB be the given finite straight line.

1.prop.1.p37

Thus it is required to construct an equilateral triangle on the straight line AB.

1.prop.1.p38

With centre A and distance AB let the circle BCD be described; [Post. 3] again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1]

1.prop.1.p39

Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15]

1.prop.1.p40

Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]

1.prop.1.p41

But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB.

1.prop.1.p42

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore CA is also equal to CB.

1.prop.1.p43

Therefore the three straight lines CA, AB, BC are equal to one another.

1.prop.1.p44

Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.

1.prop.1.p45

QED. (Being) what it was required to do.

1.prop.2.p46

Enunciation To place at a given point (as an extremity) a straight line equal to a given straight line.

1.prop.2.p47

Proof. Let A be the given point, and BC the given straight line.

1.prop.2.p48

Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC.

1.prop.2.p49

From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangle DAB be constructed. [I. 1]

1.prop.2.p50

Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let the circle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3]

1.prop.2.p51

Then, since the point B is the centre of the circle CGH, BC is equal to BG.

1.prop.2.p52

Again, since the point D is the centre of the circle GKL, DL is equal to DG.

1.prop.2.p53

And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C.N. 3]

1.prop.2.p54

But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG.

1.prop.2.p55

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore AL is also equal to BC.

1.prop.2.p56

Therefore at the given point A the straight line AL is placed equal to the given straight line BC.

1.prop.2.p57

QED. (Being) what it was required to do.

1.prop.3.p58

Enunciation Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

1.prop.3.p59

Proof. Let AB, C be the-two given unequal straight lines, and let AB be the greater of them.

1.prop.3.p60

Thus it is required to cut off from AB the greater a straight line equal to C the less.

1.prop.3.p61

At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15] But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]

1.prop.3.p62

Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.

1.prop.3.p63

QED. (Being) what it was required to do.

1.prop.4.p64

Enunciation If two triangles have the 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 remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

1.prop.4.p65

Proof. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.

1.prop.4.p66

I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

1.prop.4.p67

For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE.

1.prop.4.p68

Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF.

1.prop.4.p69

But B also coincided with E; hence the base BC will coincide with the base EF.

1.prop.4.p70

[For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4]

1.prop.4.p71

Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it.

1.prop.4.p72

And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

1.prop.4.p73

Therefore etc.

1.prop.4.p74

QED. (Being) what it was required to prove.

1.prop.5.p75

Enunciation In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

1.prop.5.p76

Proof. Let ABC be an isosceles triangle having the side AB equal to the side AC; and let the straight lines BD, CE be produced further in a straight line with AB, AC. [Post. 2]

1.prop.5.p77

I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.

1.prop.5.p78

Let a point F be taken at random on BD; from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1]

1.prop.5.p79

Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively; and they contain a common angle, the angle FAG. Therefore the base FC is equal to the base GB, and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4]

1.prop.5.p80

And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG.

1.prop.5.p81

But FC was also proved equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.

1.prop.5.p82

Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG is equal to the angle BCF, the remaining angle ABC is equal to the remaining angle ACB; and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB; and they are under the base.

1.prop.5.p83

Therefore etc.

1.prop.5.p84

Q. E. D.

1.prop.6.p85

Enunciation If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

1.prop.6.p86

Proof. Let ABC be a triangle having the angle ABC equal to the angle ACB;

1.prop.6.p87

I say that the side AB is also equal to the side AC.

1.prop.6.p88

For, if AB is unequal to AC, one of them is greater.

1.prop.6.p89

Let AB be greater; and from AB the greater let DB be cut off equal to AC the less;

1.prop.6.p90

let DC be joined.

1.prop.6.p91

Then, since DB is equal to AC, and BC is common, the two sides DB, BC are equal to the two sides AC, CB respectively; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.

1.prop.6.p92

Therefore etc.

1.prop.6.p93

Q. E. D.

1.prop.7.p94

Enunciation Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

1.prop.7.p95

Proof. For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

1.prop.7.p96

Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5] therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB.

1.prop.7.p97

Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it: which is impossible.

1.prop.7.p98

Therefore etc.

1.prop.7.p99

Q. E. D.

1.prop.8.p100

Enunciation If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

1.prop.8.p101

Proof. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let them have the base BC equal to the base EF;

1.prop.8.p102

I say that the angle BAC is also equal to the angle EDF.

1.prop.8.p103

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

1.prop.8.p104

Then, BC coinciding with EF, BA, AC will also coincide with ED, DF; for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF, then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [I. 7]

1.prop.8.p105

Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF; they will therefore coincide, so that the angle BAC will also coincide with the angle EDF, and will be equal to it.

1.prop.8.p106

If therefore etc.

1.prop.8.p107

Q. E. D.

1.prop.9.p108

Enunciation To bisect a given rectilineal angle.

1.prop.9.p109

Proof. Let the angle BAC be the given rectilineal angle.

1.prop.9.p110

Thus it is required to bisect it.

1.prop.9.p111

Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined.

1.prop.9.p112

I say that the angle BAC has been bisected by the straight line AF.

1.prop.9.p113

For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively.

1.prop.9.p114

And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. [I. 8]

1.prop.9.p115

Therefore the given rectilineal angle BAC has been bisected by the straight line AF.

1.prop.9.p116

Q. E. F.

1.prop.10.p117

Enunciation To bisect a given finite straight line.

1.prop.10.p118

Proof. Let AB be the given finite straight line.

1.prop.10.p119

Thus it is required to bisect the finite straight line AB.

1.prop.10.p120

Let the equilateral triangle ABC be constructed on it, [I. 1] and let the angle ACB be bisected by the straight line CD; [I. 9]

1.prop.10.p121

I say that the straight line AB has been bisected at the point D.

1.prop.10.p122

For, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base BD. [I. 4]

1.prop.10.p123

Therefore the given finite straight line AB has been bisected at D.

1.prop.10.p124

Q. E. F.

1.prop.11.p125

Enunciation To draw a straight line at right angles to a given straight line from a given point on it.

1.prop.11.p126

Proof. Let AB be the given straight line, and C the given point on it.

1.prop.11.p127

Thus it is required to draw from the point C a straight line at right angles to the straight line AB.

1.prop.11.p128

Let a point D be taken at random on AC; let CE be made equal to CD; [I. 3] on DE let the equilateral triangle FDE be constructed, [I. 1] and let FC be joined;

1.prop.11.p129

I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

1.prop.11.p130

For, since DC is equal to CE, and CF is common, the two sides DC, CF are equal to the two sides EC, CF respectively; and the base DF is equal to the base FE; therefore the angle DCF is equal to the angle ECF; [I. 8] and they are adjacent angles.

1.prop.11.p131

But, 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; [Def. 10] therefore each of the angles DCF, FCE is right.

1.prop.11.p132

Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.

1.prop.11.p133

Q. E. F.

1.prop.12.p134

Enunciation To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.

1.prop.12.p135

Proof. Let AB be the given infinite straight line, and C the given point which is not on it; thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

1.prop.12.p136

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3] let the straight line EG be bisected at H, [I. 10] and let the straight lines CG, CH, CE be joined. [Post. 1]

1.prop.12.p137

I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

1.prop.12.p138

For, since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE; therefore the angle CHG is equal to the angle EHC. [I. 8] And they are adjacent angles.

1.prop.12.p139

But, 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. [Def. 10]

1.prop.12.p140

Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

1.prop.12.p141

Q. E. F.

1.prop.13.p142

Enunciation If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.

1.prop.13.p143

Proof. For let any straight line AB set up on the straight line CD make the angles CBA, ABD;

1.prop.13.p144

I say that the angles CBA, ABD are either two right angles or equal to two right angles.

1.prop.13.p145

Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]

1.prop.13.p146

But, if not, let BE be drawn from the point B at right angles to CD; [I. 11] therefore the angles CBE, EBD are two right angles.

1.prop.13.p147

Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2]

1.prop.13.p148

Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each; therefore the angles DBA. ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2]

1.prop.13.p149

But the angles CBE, EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [C. N. 1] therefore the angles CBE, EBD are also equal to the angles DBA, ABC. But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.

1.prop.13.p150

Therefore etc.

1.prop.13.p151

Q. E. D.

1.prop.14.p152

Enunciation If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.

1.prop.14.p153

Proof. For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;

1.prop.14.p154

I say that BD is in a straight line with CB.

1.prop.14.p155

For, if BD is not in a straight line with BC, let BE be in a straight line with CB.

1.prop.14.p156

Then, since the straight line AB stands on the straight line CBE, the angles ABC, ABE are equal to two right angles. [I. 13] But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]

1.prop.14.p157

Let the angle CBA be subtracted from each; therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB.

1.prop.14.p158

Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD.

1.prop.14.p159

Therefore etc.

1.prop.14.p160

Q. E. D.

1.prop.15.p161

Enunciation If two straight lines cut one another, they make the vertical angles equal to one another.

1.prop.15.p162

Proof. For let the straight lines AB, CD cut one another at the point E;

1.prop.15.p163

I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED.

1.prop.15.p164

For, since the straight line AE stands on the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles [I. 13]

1.prop.15.p165

Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13]

1.prop.15.p166

But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED DEB. [Post. 4 and C. N. 1] Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3]

1.prop.15.p167

Similarly it can be proved that the angles CEB, DEA are also equal.

1.prop.15.p168

Therefore etc. Q. E. D.

1.prop.15.p169

[Porism. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.

1.prop.16.p170

Enunciation In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

1.prop.16.p171

Proof. Let ABC be a triangle, and let one side of it BC be produced to D;

1.prop.16.p172

I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.

1.prop.16.p173

Let AC be bisected at E [I. 10], and let BE be joined and produced in a straight line to F;

1.prop.16.p174

let EF be made equal to BE[I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2].

1.prop.16.p175

Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15] Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle BAE is equal to the angle ECF.

1.prop.16.p176

But the angle ECD is greater than the angle ECF; [C. N. 5] therefore the angle ACD is greater than the angle BAE.

1.prop.16.p177

Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well.

1.prop.16.p178

Therefore etc.

1.prop.16.p179

Q. E. D.

1.prop.17.p180

Enunciation In any triangle two angles taken together in any manner are less than two right angles.

1.prop.17.p181

Proof. Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles.

1.prop.17.p182

For let BC be produced to D. [Post. 2]

1.prop.17.p183

Then, since the angle ACD is an exterior angle of the triangle ABC,

1.prop.17.p184

it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA. But the angles ACD, ACB are equal to two right angles. [I. 13]

1.prop.17.p185

Therefore the angles ABC, BCA are less than two right angles.

1.prop.17.p186

Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.

1.prop.17.p187

Therefore etc.

1.prop.17.p188

Q. E. D.

1.prop.18.p189

Enunciation In any triangle the greater side subtends the greater angle.

1.prop.18.p190

Proof. For let ABC be a triangle having the side AC greater than AB;

1.prop.18.p191

I say that the angle ABC is also greater than the angle BCA.

1.prop.18.p192

For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD bejoined.

1.prop.18.p193

Then, since the angle ADB is an exterior angle of the triangle BCD,

1.prop.18.p194

it is greater than the interior and opposite angle DCB. [I. 16]

1.prop.18.p195

But the angle ADB is equal to the angle ABD, since the side AB is equal to AD; therefore the angle ABD is also greater than the angle ACB; therefore the angle ABC is much greater than the angle ACB.

1.prop.18.p196

Therefore etc.

1.prop.18.p197

Q. E. D.

1.prop.19.p198

Enunciation In any triangle the greater angle is subtended by the greater side.

1.prop.19.p199

Proof. Let ABC be a triangle having the angle ABC greater than the angle BCA;

1.prop.19.p200

I say that the side AC is also greater than the side AB.

1.prop.19.p201

For, if not, AC is either equal to AB or less.

1.prop.19.p202

Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I. 5] but it is not; therefore AC is not equal to AB.

1.prop.19.p203

Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I. 18] but it is not; therefore AC is not less than AB.

1.prop.19.p204

And it was proved that it is not equal either. Therefore AC is greater than AB.

1.prop.19.p205

Therefore etc.

1.prop.19.p206

Q. E. D.

1.prop.20.p207

Enunciation In any triangle two sides taken together in any manner are greater than the remaining one.

1.prop.20.p208

Proof. For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB.

1.prop.20.p209

For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.

1.prop.20.p210

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5] therefore the angle BCD is greater than the angle ADC. [C.N. 5]

1.prop.20.p211

And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC.

1.prop.20.p212

But DA is equal to AC; therefore BA, AC are greater than BC.

1.prop.20.p213

Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

1.prop.20.p214

Therefore etc.

1.prop.20.p215

Q. E. D.

1.prop.21.p216

Enunciation If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

1.prop.21.p217

Proof. On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle;

1.prop.21.p218

I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC.

1.prop.21.p219

For let BD be drawn through to E.

1.prop.21.p220

Then, since in any triangle two sides are greater than the remaining one, [I. 20] therefore, in the triangle ABE, the two sides AB, AE are greater than BE.

1.prop.21.p221

Let EC be added to each; therefore BA, AC are greater than BE, EC.

1.prop.21.p222

Again, since, in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each; therefore CE, EB are greater than CD, DB.

1.prop.21.p223

But BA, AC were proved greater than BE, EC; therefore BA, AC are much greater than BD, DC.

1.prop.21.p224

Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16] therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED.

1.prop.21.p225

For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB; therefore the angle BDC is much greater than the angle BAC.

1.prop.21.p226

Therefore etc.

1.prop.21.p227

Q. E. D.

1.prop.22.p228

Enunciation 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. [I. 20]

1.prop.22.p229

Proof. Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C.

1.prop.22.p230

Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3]

1.prop.22.p231

With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined;

1.prop.22.p232

I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

1.prop.22.p233

For, since the point F is the centre of the circle DKL, FD is equal to FK.

1.prop.22.p234

But FD is equal to A; therefore KF is also equal to A.

1.prop.22.p235

Again, since the point G is the centre of the circle LKH, GH is equal to GK.

1.prop.22.p236

But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C.

1.prop.22.p237

Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.

1.prop.22.p238

Q. E. F.

1.prop.23.p239

Enunciation On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.

1.prop.23.p240

Proof. Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;

1.prop.23.p241

thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE.

1.prop.23.p242

On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.

1.prop.23.p243

Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8]

1.prop.23.p244

Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.

1.prop.23.p245

Q. E. F.

1.prop.24.p246

Enunciation If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.

1.prop.24.p247

Proof. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D;

1.prop.24.p248

I say that the base BC is also greater than the base EF.

1.prop.24.p249

For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined.

1.prop.24.p250

Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG, respectively; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. [I. 4]

1.prop.24.p251

Again, since DF is equal to DG, the angle DGF is also equal to the angle DFG; [I. 5] therefore the angle DFG is greater than the angle EGF.

1.prop.24.p252

Therefore the angle EFG is much greater than the angle EGF.

1.prop.24.p253

And, since EFG is a triangle having the angle EFG greater than the angle EGF, and the greater angle is subtended by the greater side, [I. 19] the side EG is also greater than EF.

1.prop.24.p254

But EG is equal to BC. Therefore BC is also greater than EF.

1.prop.24.p255

Therefore etc.

1.prop.24.p256

Q. E. D.

1.prop.25.p257

Enunciation If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

1.prop.25.p258

Proof. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let the base BC be greater than the base EF;

1.prop.25.p259

I say that the angle BAC is also greater than the angle EDF.

1.prop.25.p260

For, if not, it is either equal to it or less.

1.prop.25.p261

Now the angle BAC is not equal to the angle EDF; for then the base BC would also have been equal to the base EF, [I. 4] but it is not; therefore the angle BAC is not equal to the angle EDF.

1.prop.25.p262

Neither again is the angle BAC less than the angle EDF; for then the base BC would also have been less than the base EF, [I. 24] but it is not; therefore the angle BAC is not less than the angle EDF.

1.prop.25.p263

But it was proved that it is not equal either; therefore the angle BAC is greater than the angle EDF.

1.prop.25.p264

Therefore etc.

1.prop.25.p265

Q. E. D.

1.prop.26.p266

Enunciation If two triangles have the 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 remaining angle.

1.prop.26.p267

Proof. Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;

1.prop.26.p268

I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF, and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF.

1.prop.26.p269

For, if AB is unequal to DE, one of them is greater.

1.prop.26.p270

Let AB be greater, and let BG be made equal to DE; and let GC be joined.

1.prop.26.p271

Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle GCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA; therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible. Therefore AB is not unequal to DE, and is therefore equal to it.

1.prop.26.p272

But BC is also equal to EF; therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4]

1.prop.26.p273

Again, let sides subtending equal angles be equal, as AB to DE;

1.prop.26.p274

I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

1.prop.26.p275

For, if BC is unequal to EF, one of them is greater.

1.prop.26.p276

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

1.prop.26.p277

Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle BHA is equal to the angle EFD.

1.prop.26.p278

But the angle EFD is equal to the angle BCA; therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA: which is impossible. [I. 16]

1.prop.26.p279

Therefore BC is not unequal to EF, and is therefore equal to it.

1.prop.26.p280

But AB is also equal to DE; therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle EDF. [I. 4]

1.prop.26.p281

Therefore etc.

1.prop.26.p282

Q. E. D.

1.prop.27.p283

Enunciation If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

1.prop.27.p284

Proof. For let the straight line EF falling on the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another;

1.prop.27.p285

I say that AB is parallel to CD.

1.prop.27.p286

For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.

1.prop.27.p287

Let them be produced and meet, in the direction of B, D, at G.

1.prop.27.p288

Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG: which is impossible. [I. 16]

1.prop.27.p289

Therefore AB, CD when produced will not meet in the direction of B, D.

1.prop.27.p290

Similarly it can be proved that neither will they meet towards A, C.

1.prop.27.p291

But straight lines which do not meet in either direction are parallel; [Def. 23] therefore AB is parallel to CD.

1.prop.27.p292

Therefore etc.

1.prop.27.p293

Q. E. D.

1.prop.28.p294

Enunciation If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

1.prop.28.p295

Proof. For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles;

1.prop.28.p296

I say that AB is parallel to CD.

1.prop.28.p297

For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH, [I. 15] the angle AGH is also equal to the angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]

1.prop.28.p298

Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13] the angles AGH, BGH are equal to the angles BGH, GHD.

1.prop.28.p299

Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]

1.prop.28.p300

Therefore etc.

1.prop.28.p301

Q. E. D.

1.prop.29.p302

Enunciation 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 interior angles on the same side equal to two right angles.

1.prop.29.p303

Proof. For let the straight line EF fall on the parallel straight lines AB, CD;

1.prop.29.p304

I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles.

1.prop.29.p305

For, if the angle AGH is unequal to the angle GHD, one of them is greater.

1.prop.29.p306

Let the angle AGH be greater.

1.prop.29.p307

Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD.

1.prop.29.p308

But the angles AGH, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are less than two right angles.

1.prop.29.p309

But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel.

1.prop.29.p310

Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it.

1.prop.29.p311

Again, the angle AGH is equal to the angle EGB; [I. 15] therefore the angle EGB is also equal to the angle GHD. [C.N. 1]

1.prop.29.p312

Let the angle BGH be added to each; therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C.N. 2]

1.prop.29.p313

But the angles EGB, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are also equal to two right angles.

1.prop.29.p314

Therefore etc.

1.prop.29.p315

Q. E. D.

1.prop.30.p316

Enunciation Straight lines parallel to the same straight line are also parallel to one another.

1.prop.30.p317

Proof. Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD.

1.prop.30.p318

For let the straight line GK fall upon them;

1.prop.30.p319

Then, since the straight line GK has fallen on the parallel straight lines AB, EF, the angle AGK is equal to the angle GHF. [I. 29]

1.prop.30.p320

Again, since the straight line GK has fallen on the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD. [I. 29]

1.prop.30.p321

But the angle AGK was also proved equal to the angle GHF; therefore the angle AGK is also equal to the angle GKD; [C.N. 1] and they are alternate.

1.prop.30.p322

Therefore AB is parallel to CD.

1.prop.30.p323

Q. E. D.

1.prop.31.p324

Enunciation Through a given point to draw a straight line parallel to a given straight line.

1.prop.31.p325

Proof. Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC.

1.prop.31.p326

Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23]; and let the straight line AF be produced in a straight line with EA.

1.prop.31.p327

Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27]

1.prop.31.p328

Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC.

1.prop.31.p329

Q. E. F.

1.prop.32.p330

Enunciation In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

1.prop.32.p331

Proof. Let ABC be a triangle, and let one side of it BC be produced to D;

1.prop.32.p332

I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles.

1.prop.32.p333

For let CE be drawn through the point C parallel to the straight line AB. [I. 31]

1.prop.32.p334

Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29]

1.prop.32.p335

Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29]

1.prop.32.p336

But the angle ACE was also proved equal to the angle BAC; therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC.

1.prop.32.p337

Let the angle ACB be added to each; therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB.

1.prop.32.p338

But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles.

1.prop.32.p339

Therefore etc.

1.prop.32.p340

Q. E. D.

1.prop.33.p341

Enunciation The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

1.prop.33.p342

Proof. Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel.

1.prop.33.p343

Let BC be joined.

1.prop.33.p344

Then, since AB is parallel to CD, and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

1.prop.33.p345

And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the griangle ABC is equal to the triangle DCB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle ACB is equal to the angle CBD.

1.prop.33.p346

And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another, AC is parallel to BD. [I. 27]

1.prop.33.p347

And it was also proved equal to it.

1.prop.33.p348

Therefore etc.

1.prop.33.p349

Q. E. D.

1.prop.34.p350

Enunciation In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

1.prop.34.p351

Proof. Let ACDB be a parallelogrammic area, and BC its diameter; I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

1.prop.34.p352

For, since AB is parallel to CD, and the straight line BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]

1.prop.34.p353

Again, since AC is parallel to BD, and BChas fallen upon them, the alternate angles ACB, CBD are equal to one another. [I. 29]

1.prop.34.p354

Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26] therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB.

1.prop.34.p355

And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C.N. 2] And the angle BAC was also proved equal to the angle CDB.

1.prop.34.p356

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

1.prop.34.p357

I say, next, that the diameter also bisects the areas.

1.prop.34.p358

For, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and the angle ABC is equal to the angle BCD; therefore the base AC is also equal to DB, and the triangle ABC is equal to the triangle DCB. [I. 4]

1.prop.34.p359

Therefore the diameter BC bisects the parallelogram ACDB.

1.prop.34.p360

Q. E. D.

1.prop.35.p361

Enunciation Parallelograms which are on the same base and in the same parallels are equal to one another.

1.prop.35.p362

Proof. Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC; I say that ABCD is equal to the parallelogram EBCF.

1.prop.35.p363

For, since ABCD is a parallelogram, AD is equal to BC. [I. 34]

1.prop.35.p364

For the same reason also EF is equal to BC, so that AD is also equal to EF; [C.N. 1] and DE is common; therefore the whole AE is equal to the whole DF. [C.N. 2]

1.prop.35.p365

But AB is also equal to DC; [I. 34] therefore the two sides EA, AB are equal to the two sides FD, DC respectively, and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29] therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4]

1.prop.35.p366

Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

1.prop.35.p367

Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

1.prop.35.p368

Therefore etc.

1.prop.35.p369

Q. E. D.

1.prop.36.p370

Enunciation Parallelograms which are on equal bases and in the same parallels are equal to one another.

1.prop.36.p371

Proof. Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG; I say that the parallelogram ABCD is equal to EFGH.

1.prop.36.p372

For let BE, CH be joined.

1.prop.36.p373

Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C.N. 1]

1.prop.36.p374

But they are also parallel.

1.prop.36.p375

And EB, HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33]

1.prop.36.p376

Therefore EBCH is a parallelogram. [I. 34]

1.prop.36.p377

And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35]

1.prop.36.p378

For the same reason also EFGH is equal to the same EBCH; [I. 35] so that the parallelogram ABCD is also equal to EFGH. [C.N. 1]

1.prop.36.p379

Therefore etc. Q. E. D.

1.prop.37.p380

Enunciation Triangles which are on the same base and in the same parallels are equal to one another.

1.prop.37.p381

Proof. Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC; I say that the triangle ABC is equal to the triangle DBC.

1.prop.37.p382

Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I. 31] and through C let CF be drawn parallel to BD. [I. 31]

1.prop.37.p383

Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,

1.prop.37.p384

for they are on the same base BC and in the same parallels BC, EF. [I. 35]

1.prop.37.p385

Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]

1.prop.37.p386

And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]

1.prop.37.p387

[But the halves of equal things are equal to one another.]

1.prop.37.p388

Therefore the triangle ABC is equal to the triangle DBC.

1.prop.37.p389

Therefore etc.

1.prop.37.p390

Q. E. D.

1.prop.38.p391

Enunciation Triangles which are on equal bases and in the same parallels are equal to one another.

1.prop.38.p392

Proof. Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF.

1.prop.38.p393

For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.

1.prop.38.p394

Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;

1.prop.38.p395

for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]

1.prop.38.p396

Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]

1.prop.38.p397

And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]

1.prop.38.p398

[But the halves of equal things are equal to one another.]

1.prop.38.p399

Therefore the triangle ABC is equal to the triangle DEF.

1.prop.38.p400

Therefore etc.

1.prop.38.p401

Q. E. D.

1.prop.39.p402

Enunciation Equal triangles which are on the same base and on the same side are also in the same parallels.

1.prop.39.p403

Proof. Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it; [I say that they are also in the same parallels.]

1.prop.39.p404

And [For] let AD be joined; I say that AD is parallel to BC.

1.prop.39.p405

For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.

1.prop.39.p406

Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels. [I. 37]

1.prop.39.p407

But ABC is equal to DBC; therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.

1.prop.39.p408

Therefore AE is not parallel to BC.

1.prop.39.p409

Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC.

1.prop.39.p410

Therefore etc.

1.prop.39.p411

Q. E. D.

1.prop.40.p412

Equal triangles which are on equal bases and on the same side are also in the same parallels.

1.prop.40.p413

Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.

1.prop.40.p414

I say that they are also in the same parallels.

1.prop.40.p415

For let AD be joined; I say that AD is parallel to BE.

1.prop.40.p416

For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.

1.prop.40.p417

Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]

1.prop.40.p418

But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible. Therefore AF is not parallel to BE.

1.prop.40.p419

Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.

1.prop.40.p420

Therefore etc. Q. E. D.]

1.prop.41.p421

Enunciation If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

1.prop.41.p422

Proof. For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;

1.prop.41.p423

I say that the parallelogram ABCD is double of the triangle BEC.

1.prop.41.p424

For let AC be joined.

1.prop.41.p425

Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I. 37]

1.prop.41.p426

But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; [I. 34] so that the parallelogram ABCD is also double of the triangle EBC.

1.prop.41.p427

Therefore etc.

1.prop.41.p428

Q. E. D.

1.prop.42.p429

Enunciation To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

1.prop.42.p430

Proof. Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC.

1.prop.42.p431

Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.

1.prop.42.p432

Then FECG is a parallelogram.

1.prop.42.p433

And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38] therefore the triangle ABC is double of the triangle AEC.

1.prop.42.p434

But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC.

1.prop.42.p435

And it has the angle CEF equal to the given angle D.

1.prop.42.p436

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q. E. F.

1.prop.43.p437

Enunciation In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

1.prop.43.p438

Proof. Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;

1.prop.43.p439

I say that the complement BK is equal to the complement KD.

1.prop.43.p440

For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34]

1.prop.43.p441

Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. For the same reason the triangle KFC is also equal to KGC.

1.prop.43.p442

Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]

1.prop.43.p443

And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]

1.prop.43.p444

Therefore etc.

1.prop.43.p445

Q. E. D.

1.prop.44.p446

Enunciation To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

1.prop.44.p447

Proof. Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

1.prop.44.p448

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42]; let it be placed so that BE is in a straight line with AB; letFG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31]

1.prop.44.p449

Let HB be joined.

1.prop.44.p450

Then, since the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are equal to two right angles. [I. 29] Therefore the angles BHG, GFE are less than two right angles; and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5] therefore HB, FE, when produced, will meet.

1.prop.44.p451

Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31] and let HA, GB be produced to the points L, M.

1.prop.44.p452

Then HLKF is a parallelogram, HK is its diameter, and AG, ME are parallelograms. and LB, BF the so-called complements, about HK; therefore LB is equal to BF. [I. 43]

1.prop.44.p453

But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]

1.prop.44.p454

And, since the angle GBE is equal to the angle ABM, [I. 15] while the angle GBE is equal to D, the angle ABM is also equal to the angle D.

1.prop.44.p455

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D.

1.prop.44.p456

Q. E. F.

1.prop.45.p457

Enunciation To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

1.prop.45.p458

Proof. Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

1.prop.45.p459

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44]

1.prop.45.p460

Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1]

1.prop.45.p461

Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.

1.prop.45.p462

But the angles FKH, KHG are equal to two right angles; [I. 29] therefore the angles KHG, GHM are also equal to two right angles.

1.prop.45.p463

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles; therefore KH is in a straight line with HM. [I. 14]

1.prop.45.p464

And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]

1.prop.45.p465

Let the angle HGL be added to each; therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2]

1.prop.45.p466

But the angles MHG, HGL are equal to two right angles; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1] Therefore FG is in a straight line with GL. [I. 14]

1.prop.45.p467

And, since FK is equal and parallel to HG, [I. 34] and HG to ML also, KF is also equal and parallel to ML; [C.N. 1; I. 30] and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel. [I. 33] Therefore KFLM is a parallelogram.

1.prop.45.p468

And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

1.prop.45.p469

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.

1.prop.45.p470

Q. E. F.

1.prop.46.p471

Enunciation On a given straight line to describe a square.

1.prop.46.p472

Proof. Let AB be the given straight line; thus it is required to describe a square on the straight line AB.

1.prop.46.p473

Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]

1.prop.46.p474

Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34]

1.prop.46.p475

But AB is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral.

1.prop.46.p476

I say next that it is also right-angled.

1.prop.46.p477

For, since the straight line AD falls upon the parallels AB, DE, the angles BAD, ADE are equal to two right angles. [I. 29]

1.prop.46.p478

But the angle BAD is right; therefore the angle ADE is also right.

1.prop.46.p479

And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled.

1.prop.46.p480

And it was also proved equilateral.

1.prop.46.p481

Therefore it is a square; and it is described on the straight line AB.

1.prop.46.p482

Q. E. F.

1.prop.47.p483

Enunciation In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

1.prop.47.p484

Proof. Let ABC be a right-angled triangle having the angle BAC right;

1.prop.47.p485

I say that the square on BC is equal to the squares on BA, AC.

1.prop.47.p486

For let there be described on BC the square BDEC, and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined.

1.prop.47.p487

Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC, AG not lying on the same side make the adjacent angles equal to two right angles; therefore CA is in a straight line with AG. [I. 14]

1.prop.47.p488

For the same reason BA is also in a straight line with AH.

1.prop.47.p489

And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each; therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]

1.prop.47.p490

And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively, and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I. 4]

1.prop.47.p491

Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL. [I. 41]

1.prop.47.p492

And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41]

1.prop.47.p493

[But the doubles of equals are equal to one another.] Therefore the parallelogram BL is also equal to the square GB.

1.prop.47.p494

Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC; therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]

1.prop.47.p495

And the square BDEC is described on BC, and the squares GB, HC on BA, AC.

1.prop.47.p496

Therefore the square on the side BC is equal to the squares on the sides BA, AC.

1.prop.47.p497

Therefore etc.

1.prop.47.p498

Q. E. D.

1.prop.48.p499

Enunciation If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

1.prop.48.p500

Proof. For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC;

1.prop.48.p501

I say that the angle BAC is right.

1.prop.48.p502

For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined.

1.prop.48.p503

Since DA is equal to AB, the square on DA is also equal to the square on AB.

1.prop.48.p504

Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC.

1.prop.48.p505

But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC.

1.prop.48.p506

And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] But the angle DAC is right; therefore the angle BAC is also right.

1.prop.48.p507

Therefore etc.

1.prop.48.p508

Q. E. D.

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