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Euclids Elements

A Modern Treatment

Roger Bickley

Book 1

2

3

Contents Introduction ............................................................................................................................................ 9

Note To Symbols Used ......................................................................................................................... 12

Definitions ............................................................................................................................................ 12

Postulates ............................................................................................................................................. 14

Postulate 1 ........................................................................................................................................ 14

Postulate 2 ........................................................................................................................................ 14

Postulate 3 ........................................................................................................................................ 14

Postulate 4 ........................................................................................................................................ 14

Postulate 5 ........................................................................................................................................ 14

Common Notions.................................................................................................................................. 14

Common Notion 1 ............................................................................................................................ 14

Common Notions 2,3 ........................................................................................................................ 14

Common Notion 4 ............................................................................................................................ 14

Common Notion 5 ............................................................................................................................ 14

Proposition 1 ........................................................................................................................................ 15

On a given finite straight line to construct an equilateral ............................................................ 15

Proposition 2 ........................................................................................................................................ 16

To place at a given point (as an extremity) a straight line equal to a given straight line ................ 16

Proposition 3 ........................................................................................................................................ 18

Given two unequal straight lines, to cut off from the greater a straight line equal to the less ....... 18

Proposition 4 ........................................................................................................................................ 19

If two s 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 will be equal to the ,

and the remaining angle will be equal to the remaining angles respectively, namely those which the

equal sides subtend. ......................................................................................................................... 19

Proposition 5 ........................................................................................................................................ 21

In isosceles s 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. .................................. 21

Proposition 6 ........................................................................................................................................ 23

4

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

equal to one another ........................................................................................................................ 23

Proposition 7 ........................................................................................................................................ 24

Given two straight line 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 .......................................................................... 24

Proposition 8 ........................................................................................................................................ 25

If two s 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 line ...................... 25

Proposition 9 ........................................................................................................................................ 27

To bisect a given rectilineal angle .................................................................................................... 27

Proposition 10 ...................................................................................................................................... 28

To bisect a given finite straight line.................................................................................................. 28

Proposition 11 ...................................................................................................................................... 29

To draw a straight line at right angles to a given straight line from a given point on it .................. 29

Proposition 12 ...................................................................................................................................... 30

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

line .................................................................................................................................................... 30

Proposition 13 ...................................................................................................................................... 31

If a straight line set up on a straight line makes angles, it will make either two right angles or angles

equal to two right angles .................................................................................................................. 31

Proposition 14 ...................................................................................................................................... 32

If with any straight line, and at a point on it, two straight line 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 ............................................................................................................................................. 32

Proposition 15 ...................................................................................................................................... 33

If two straight lines cut one another, they make the vertical angles equal to each other. ............. 33

Proposition 16 ...................................................................................................................................... 34

In any , if one of the sides be produced, the exterior angle is greater than either of the interior and

opposite angles. ............................................................................................................................... 34

Proposition 17 ...................................................................................................................................... 36

In any two angles taken together in any manner are less than two right angles. ........................ 36

5

Proposition 18 ...................................................................................................................................... 37

In any the greater side subtends the greater angle. ..................................................................... 37

Proposition 19 ...................................................................................................................................... 38

In any the greater angle is subtended by the greater side. .......................................................... 38

Proposition 20 ...................................................................................................................................... 39

In any two sides taken together in any manner are greater than the remaining one. ................. 39

Proposition 21 ...................................................................................................................................... 40

If on one of the sides of a , from its extremities, there be constructed two straight line meeting within

the , the straight lines so constructed will be less than the remaining two sides of the , but will

contain a greater angle..................................................................................................................... 40

Proposition 22 ...................................................................................................................................... 42

Out of three straight lines, which are equal to three given straight lines, to construct a : thus it is

necessary that two of the straight lines taken together in any manner should be greater than the

remaining one. ................................................................................................................................. 42

Proposition 23 ...................................................................................................................................... 44

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

angle. ................................................................................................................................................ 44

Proposition 24 ...................................................................................................................................... 45

If two s 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.

.......................................................................................................................................................... 45

Proposition 25 ...................................................................................................................................... 47

If two s 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.

.......................................................................................................................................................... 47

Proposition 26 ...................................................................................................................................... 49

If two s 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. ............................................................................................................................... 49

Proposition 27 ...................................................................................................................................... 52

If a straight line falling on two straight lines makes the alternate angles equal to one another, the

straight lines will be parallel to one another.................................................................................... 52

6

Proposition 28 ...................................................................................................................................... 53

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 of the same side equal to two right angles, the straight

lines will be parallel to one another. ................................................................................................ 53

Proposition 29 ...................................................................................................................................... 55

A straight line falling on parallel 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. ...................................................................................................................................... 55

Proposition 30 ...................................................................................................................................... 57

Straight lines parallel to the same straight line are also parallel to one another. ........................... 57

Proposition 31 ...................................................................................................................................... 58

Through a given point to draw a straight line parallel to a given straight line. ............................... 58

Proposition 32 ...................................................................................................................................... 59

In any , 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 are equal to two right angles. ................................. 59

Proposition 33 ...................................................................................................................................... 61

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. ................................................... 61

Proposition 34 ...................................................................................................................................... 62

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

bisects the areas. .............................................................................................................................. 62

Proposition 35 ...................................................................................................................................... 64

Parallelograms which on the same base and in the same parallels are equal to one another ....... 64

Proposition 36 ...................................................................................................................................... 65

Parallelograms which are on equal bases and in the same parallels are equal to one another ...... 65

Proposition 37 ...................................................................................................................................... 66

s which are on the same base and in the same parallels are equal to one another. .................... 66

Proposition 38 ...................................................................................................................................... 67

s which are on equal bases and in the same parallels are equal to one another. ......................... 67

Proposition 39 ...................................................................................................................................... 68

Equal triangles which are on the same base and on the same side are also in the same parallels. 68

Proposition 40 ...................................................................................................................................... 69

7

Equal s which are on equal bases and on the same side are also in the same parallels. .............. 69

Proposition 41 ...................................................................................................................................... 70

If a parallelogram have the same base with a and be in the same parallels, the parallelogram is

double of the . ................................................................................................................................ 70

Proposition 42 ...................................................................................................................................... 71

To construct, in a given rectilineal angle, a parallelogram equal to a given . ................................ 71

Proposition 43 ...................................................................................................................................... 72

In any parallelogram the complements of the parallelogram about the diameter are equal to one

another. ............................................................................................................................................ 72

Proposition 44 ...................................................................................................................................... 73

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given . . 73

Proposition 45 ...................................................................................................................................... 75

To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure. ........ 75

Proposition 46 ...................................................................................................................................... 77

On a given straight line to describe a square. .................................................................................. 77

Proposition 47 ...................................................................................................................................... 79

In right-angled s the square on the side subtending the right angle is equal to the squares on the side

containing the right angle ................................................................................................................ 79

Proposition 48 ...................................................................................................................................... 81

If in a the square on one of the sides be equal to the square on the remaining side of the the angle

contained by the remaining two sides of the is right. ................................................................... 81

Glossary of Terms Used in the Elements .............................................................................................. 83

8

9

Introduction Euclids Elements is, perhaps, the single most significant work on Mathematics. It seeks to place

the foundations of geometry in an axiomatic and logical framework.

Starting from a set of axioms it builds a series of conclusions, each conclusion being based

on/supported by, conclusions which have been proved earlier in the system..

This book, whilst remaining true to the spirit and structure of the original, seeks to present in

everyday (mathematics) language the propositions, as described by Euclid.

A problem which has to be faced when reading a book such as the Elements is that of language.

The original language of the books was Greek, but it has been translated a number of times

including into English.

The present book takes, as its starting point, the translation by Sir Thomas L Heath.

Even though the books have been translated into English, this doesnt completely resolve the

difficulty of language words in Greek dont always have an exact translation into English and

so the translator has to use what language he feels serves the original text best.

For example terms like rectilineal angle are used where we would probably just use the word

angle.

Also, the wording of the propositions can seem to be complicated, even verbose, and certainly

not always clear to the modern reader.

I felt that, to do justice to the original text, these two issues need to be addressed to make such

a wonderful book more accessible to the modern reader. To that end I have adopted the

following strategy:

1. The propositions are stated in exactly the wording of the Euclids Elements books of Sir

Thomas Heath

2. Underneath each original proposition wording is my interpretation of the meaning of

the proposition, in modern language

3. Words which may not be familiar to modern readers, or familiar words used in an

unfamiliar way, are listed in a Glossary at the back of the book

A key feature of the Euclid Propositions is, when the proposition is of the construction form

e.g. to bisect an angle there are not only details of how to carry out the construction, but also

a proof of why the construction works!

Since Mathematics is a hierarchical discipline with results depending on previously proved

theorems/conjectures, some of the propositions (in the case of later ones) have references to

previous propositions which have proved intermediate results. In the cases of the very early

10

11

propositions, these proofs start from a set of definitions (statements which Euclid assumes to be

true, in order to build his propositional system).

The presentation retains the order of propositions in the original Elements, including references

to Propositions already proved.

12

Note To Symbols Used

In the original text (and translations) word are used to describe relationships and to signpost the

argument. In this treatment I have used some common symbols to replace the words hoping

that this will make the text less cluttered and easier to read:

identical to sometimes called congruence

not equal to

to denote a , rather than an angle thus ABC means the ABC whilst the angle ABC would

be written ABC

is put before an angle

therefore

> greater than

< less than

Q.E.D Quad est demonstratum (which was to be demonstrated) is perhaps an old fashioned way of terminating a proof (the being a more modern equivalent) but it is retained here for

historical/romantic reasons!)

Definitions Before we can begin to construct any argument we need to be clear about what we are talking

about to the extent of defining what we mean by certain objects or ideas.

Euclid goes to a great deal of trouble to explain what he means by the various objects and ideas

he uses to prove his propositions some of which, to our eyes, may seem superfluous or

unnecessarily verbose or complex. However, for completeness they are included here, and I will

leave it to the reader to interpret the definitions in the light of their own experience and/or

needs:

1. A line is that which has no part

2. A line is breadthless length

3. The extremities of a line are points

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

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

13

6. The extremities of a surface are lines

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

8. 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

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

10. When a straight line set up on a straight line makes the adjacent angles equal to one

another, each of the equal angles is right, and the straight line standing on the other is

called a perpendicular to that on which it stands

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

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

13. A boundary is that which is an extremity of anything

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

15. 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.

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

17. 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

18. 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

19. Rectilineal figures are those which are contained by straight line, trilateral figures being

those contained by three, quadrilateral those contained by four and multiplateral those

contained by more than four straight line

20. Of trilateral figures, an equilateral is that which has three sides equal, an isosceles

that which has two of its sides alone equal, and a scalene that which has its three sides

unequal

21. Further, of trilateral [three-sided] figures , a right-angled is that which has a right

angle, an obtuse-angled that which has an obtuse angle and an acute-angled that

which has its three angles acute.

22. Of quadrilateral figures, a square is that which is both equilateral and right-angled; an

oblong that which is right-angle 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.

14

23. Parallel 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.

Postulates

Postulate 1

We can draw a straight line from any point to any point

Postulate 2

We can extend a finite straight line continuously in a straight line

Postulate 3

We can draw a circle with any centre and radius

Postulate 4

All right angles are equal to one another

Postulate 5

If a straight line cuts two straight lines such that the interior angles on one side are less than

two right angles (total of 1800) the two straight lines, if we were to produce them indefinitely

meet on the same side as the two angles which, together, add up to less than 1800

Common Notions

Common Notion 1

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

Common Notions 2,3

2. If equals be added to equals, the sums are equal

3. If equals be subtracted from equals, the results are equal

Common Notion 4

Things which coincide with one another are equal to one another

Common Notion 5

The whole is greater than the part

15

Proposition 1

On a given finite straight line to construct an equilateral

[Translation: this should be clear draw any straight line and then draw an equilateral on it

with the lengths of the sides the same as the length of the line you have been given!]

This is a relatively simple construction needing a pencil, compass and ruler.

The Construction

Draw a line AB of any length (if you are drawing on a normal maths book page a 5cm line

will do)

Put the compass point on the point B and open the compasses until the pencil is on the

point A

Draw a circle so that it goes through the point A (this is the circle E)

Put the compass point on the point A and open the compasses until the pencil is on the

point B

Draw a circle so that it goes through the point B (this is the circle D)

Draw a line from A to where the circles cross (C) and from B to where the circles cross (C)

to get a ABC

The ABC is equilateral (all sides the same length and all angles equal)

Proof

Since the point A is the centre of the circle AC = AB Definition 15

Again, since B is the centre of the circle CAE BC = BA Definition 15

But we know (above) that CA = AB

Each of the straight lines CA and CB is equal to AB

And things which are equal to the same thing are also equal to one another Common Notion 1

CA is also equal to CB

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

the ABC is equilateral and it has been constructed on the given finite straight line AB

Q.E.F.

A B E D A B

C

E D

C

16

Proposition 2

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

straight line

{Translation: Given a point (A) and a straight line (BC), draw a line from the point (A) the same

length as the line you have been given]

Construction

Starting From .....

Mark any point A on the paper and draw a line BC

To Construct

We need to draw, from A a straight line equal to the given line BC

Construction

From the point A to the point B draw a straight line AB Postulate 1

On this line (AB) draw an equilateral triangle Proposition 1

Draw the lines AE and BF as extensions of the lines DA and DB

With centre D and distance DG draw the circle GKL Postulate 3

Then, since the point B is the centre of the circle CGH:

BC = BG

Again, since the point D is the centre of the circle GKL:

DL = DG

And we know that DA = DB

. A A B

C

. . A B

D

C

E

F

G

K

L

B

C

17

Therefore the remainder AL (of the line) is equal to the remainder BG (of the other line)

Common Notion 3

But we have proved that that BC = BG

each of the straight lines AL and BC is equal to BG

And tings which are equal to the same thing are also equal to one another Common Notion 1

AL = BC

Therefore at the given point A the straight line AL is equal to the line we were given BC

Q.E.F

18

Proposition 3

Given two unequal straight lines, to cut off from the greater a straight line

equal to the less

[Translation: You are given two lines of unequal length; mark off, on the larger one, a length

equal to the length of the smaller line]

Construction

1. We have two lines which we will call AB and C where AB is larger than C and we need to

cut off a length of AB equal to the length of C

2. First of all draw a line AD from A (at any angle) and equal in length to the length of the

shorter line C Proposition 2

3. Now put the point of a compass on the point A and draw a circle with radius AD which

will cut the line AB at E Postulate 3

The line AE is equal to the line C and we have achieved the construction

Proof

The point A is the centre of the circle DEF, so

AE = AD Definition 15

But C = AD

each of the straight lines AE and C is equal to AD:

So, AE is also = C Common Notion 1

given the two straight lines AB and C:

Form AB the greater length AE has been cut off and this is equal to C, the lesser length

Q.E.F.

A B

D

A B

D

E A B

C

19

Proposition 4

If two s 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 will be equal to the , and the remaining angle will be equal to the

remaining angles respectively, namely those which the equal sides subtend.

[Translation: You are given two s which have two sides equal, and the angle between the two

sides equal; then, the bases of the two s will be equal as will corresponding angles in the two

s in other words the two s will be congruent]

We have:

1. Two s ABC and DEF in which AB = DE and AC = DF

2. The angle BAC = angle EDF

To Prove

BC = EF

ABC DEF (the symbol means identical to in every respect i.e, in this case, you could pick

the one up and it would fit exactly onto the other one)

Corresponding angles in each of the s are equal to each other (i.e. angle ABC = angle DEF and

angle ACB = angle DFE

Proof

If the ABC is placed onto the DEF so that A and D are on each other, then

The point E will coincide with the point E because AB = DE

Similarly, AC will 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 equal to DF

But B coincided with E, so the base BC will coincide with the base EF

[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

And the remaining angles will also coincide with the remaining angles and be equal to them:

The angle ABC to the angle DEF, and

A

B C

D

E F

20

The angle ACB to the angle DFE

Therefore ABCDEF

Q.E.D.

21

Proposition 5

In isosceles s 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.

[an isosceles has two sides equal this proposition says that the angles at the end of the two

equal lines are also equal and, when you extend the two equal sides the two external angles

are also equal to one another]

We have:

An Isosceles in which AB = AC

To Prove

Angle BCE = angle CBD

Proof

Extend the line AB to become AD and the line AC to become AE - Postulate 2

Mark a point F anywhere on BD

Mark a point G on CE equal to AF Proposition 3

Join the points FC and GB Postulate 1

Since AF = AG (sides of an isosceles triangle) and AB = AC (we have just constructed them so)

FA = GA and AC = AB

And the sides FA = GA and AC = AB

And they contain a common angle FAG

FC = GB

A

B C

A

B C

D E

F G

22

And the AFC = AGB

And the remaining angles in the triangles will be equal to each other, respectively (i.e. as to the

sides they subtend) i.e.

ACF = ABG

And AFC = AGB

And, since the whole of AF = the whole of AG:

And AB = AC

The remainder BF is equal to the remainder CG

But, FC was also proved to be equal to GB

BF = CG and FC = GB

And BFC = CGB

While the line BC is common

the BFC is also equal to the CGB

And the remaining angles will be equal to the remaining angles respectively i.e. the angles

which the equal sides subtend

the FBC = GCB

Accordingly, since the whole angle ABG was proved equal to ACF

And in these the ABC is equal to the remaining ACB

And they are at the base of the ABC

But the FBC was also proved equal to the GCB and they are under the base

Q.E.D.

23

Proposition 6

If in a two angles be equal to one another, the sides which subtend the

equal angles will also be equal to one another

[Translation: This is like a reverse of the previous proposition if you have a with two angles

equal, then the sides opposite these angles will also be equal in other words it will be an

isosceles ]

We have angle ABC = angle ACB

To Prove

AB = AC

Proof

Let us assume that the two sides (AB and AC) are NOT equal then one of them must be bigger

than the other

Suppose AB is bigger than AC

Then we can draw a line CD such that BD = AC (because AC is smaller than AB)

We now have:

BD = AC (by construction)

BC is a side common to both s (DBC and ACB)

Angle DBC = angle ACB (base angles of an isosceles )

Therefore the base DC is equal to the base AB, and

DBC ACB (the smaller is the same as the larger one!)

Which is clearly silly!

But we assumed that AB was NOT equal to AC and this assumption got us this absurd results!

So, we must have been wrong and AB must equal AC

Q.E.D.

A

B C

A

B C

D

24

Proposition 7

Given two straight line 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

[Translation Given a ABC with a base BC, you cannot draw, on the base BC, another which

is equal in area to ABC, and IS NOT ABC i.e, this new must coincide with ABC

What we have:

A line BC and two lines drawn from wither end (BA and CA) which meet in the point A

To Prove

You cannot draw another , with the same area and on the same base unless it is on top of

the original (ABC)

Proof

First of all draw two additional straight lines from B and C respectively, to meet at a new point

D. These two lines are to be equal to the sides of the original -= i.e. CA = DA and CB = DB

Join CD

Since AC = AD (we have drawn it to be so)

Then the ACD = ADC Proposition 5

the ADC > DCB

the CDB is much greater than DCB

Since CB = DB:

The CDB = DCB

But is was also proved much greater than it! which is impossible

C

A B

C

A B

D

25

Proposition 8

If two s 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 line

[Translation: If you have two s ABC and DEF with AB = DE and AC = DF and the bases equal

i.e. BC = EF; then the BAC = EDF

We have ....

Two triangles ABC and DEF with the sides AB = DE and AC = DF, and

The bases BC = EF

To Prove

BAC = EDF

Proof

We can place the ABC onto the DEF so that the point B is placed on the point E and the

straight line BC onto the straight line EF

Then the point C will also coincide with F, because BC = EF

Then if BC coincides with EF:

BA will coincide with ED and AC coincide with DF because, if the base BC coincides with the base

EF, and the side BA coincides with ED and AC coincides with DF but fall outside the triangle as

EG and GF:

Then, given two straight lines constructed on a straight line (lines extended) and meeting at a

point there will have been constructed on the same (extended) straight line, two other straight

lines (EG and FG) meeting in another point (G) and equal to the original lines (ED an FD) but

we cannot construct such lines,

it is NOT possible that, if the base BC is on the base , the sides BA and ,AC should not coincide

with the sides ED and EF

C

A

B

D

E

F

G

26

Therefore they WILL coincide

So, the BAC WILL also coincide with EDF and will be equal to it

Q.E.D.

27

Proposition 9

To bisect a given rectilineal angle

[Translation: Cut any angle exactly in half so that there are two angles of equal size]

Construction

Mark any point D on the line AB

Mark another point E on AC such that AD = AE

Join DE

Construct an equilateral on DE DEF

Join AF

To Prove

The angle BAC has been bisected by the line AF

Proof

AD = AE (by construction)

AF is common to both s

The two sides DA and AT are equal to the two sides EA and AF respectively

The base DF is equal to the base EF

Therefore the angle DAF is equal to the angle EAF Proposition 8

Therefore the angle BAC has been bisected by the line AF

Q.E.F.

A

D E

B C F

A

B C

28

Proposition 10

To bisect a given finite straight line

[Translation: Cut a given straight line in half so that the two parts of the line are equal]

Construction

Given the straight line AB construct an equilateral on it Proposition 1

Bisect the angle ACB by the line CD Proposition 9

To Prove

The line CD bisects the line AB

Proof

AC = CB (sides of an equilateral )

The two sides AC, CD are equal to the two sides BC and CD respectively, and

The angle ACD is equal to the angle BCD Proposition 4

Therefore the line AB has been bisected at D

Q.E.F

A

B

B A

B

B

C

B

D

B

29

Proposition 11

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

[Translation: From a point on a line draw a line at 900 to the line pr perpendicular as it is als

oknown]

Construction

Draw the line AB and mark any point C on it

Mark any point D on the line AB

Mark another point E s that DE = CD

Using DE as the base draw an equilateral Proposition 1

Draw the line FC

To Prove

The line FC is perpendicular to the line AB

Proof

DC = CE (Construction)

CF is common to both s FDC and FCE

DC = EC and CF = CF and the line DF = line FE

DCF = ECF Proposition 8

And they are adjacent angles

But, when a straight line drawn on a straight line makes the adjacent angles equal to one

another each of the angles is a right-angle Definition 10

DCF and FCE are both right angles

the straight line CF has been drawn at right angles to the given line AB form the given point C

on it

Q.E.F.

A

B

B C

B

D

B

E

B

. . . . . . AB

B C

B

D

B

E

B

F

B

30

Proposition 12

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

perpendicular straight line

*So youve got a straight line and a point which isnt on the line draw a line from the point to

the line which is at right-angles to the line]

Construction

Mark any point D which is on the other side of the line from C

Put the point of compasses on the point C and draw a circle with radius CD (NOTE: it cuts

the straight line at G and E)

Bisect the straight line EG at H Proposition 10

Draw the straight lines CG, CH and CE

To Prove

CH is perpendicular to the straight line AB

Proof

GH = HE (we have just bisected it by construction)

HC is common

CA = CB radii of the same circle)

So, CHG CHE

Therefore the angle CHG = angle CHE

Since the two angles are equal and equal to 1800 (angles in a straight line) they must both be

equal to 900

Therefore CH is perpendicular to AB

Q.E.D

. C

B

B

B

A

B . DB

. C

B

E

B

G

B

H

B

B

B

A

B

31

Proposition 13

If a straight line set up on a straight line makes angles, it will make either two right

angles or angles equal to two right angles

[Translation: If you draw a straight line, and then draw another straight line onto the first line

you will end up with EITHER two right angles or two angles which add up to two right angles]

Construction

Draw a straight line DB

Pick any point on DB and draw a straight line AB to make the angles ABD and CBA

To Prove

That, either the two angles ABD and CBA are both right angles, or are EQUAL to two right angles

Proof

Suppose, to start with, that angle CBA = angle ABD

Then they are two right angles Definition 10

But, if they are NOT equal, draw a line BE at right angles to CD Proposition 11 then the angles CBE and EBD are both right angles

Then angle CBE is equal to the sum of the angles CBA and ABE

Let angle EBD be added to each

So, we have angles CBE + EBD are equal to angles CBA + ABE + EBD

And, we know that angle DBA is equal to angle DBE + angle EBA

Now let the angle ABC be added to this sum

We have angles DBA + ABC = angles DBE + EBA + ABC

But the angles CBE + EBD were proved equal to the same three angles, and things which are

equal to the same ting are also equal to each other

Therefore the angle CBE+ EBD = angles DBA + ABC

But the angles CBE and EBD add up to two right angles

Therefore the angles DBA + ABC are also equal to two right angles

D

B

C D

B

C B

B

A

B

E

B

Q.E.D

32

Proposition 14

If with any straight line, and at a point on it, two straight line 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

[Translation: Draw a straight line and go to one end of it and draw two straight lines- one in

each direction. If the angles which these lines make with the original line add up to two right

angles (1800) then the two lines make a straight line]

Construction

Draw any straight line AB

Draw two lines BC and BD, one on either side of the point B

When drawing the lines BC and BD make it so that angle ABC + angle ABD = 2 right angles

To prove

DBC is a straight line

Proof

Suppose BD is NOT in straight line with CB

Draw a line BE and let BE be in a straight line with CB

Then, since the straight line AB stands on the straight line CBE (weve just drawn it to be so)

the angles ABC + ABE = two right angles - Proposition 13

But, the angles ABC + ABD are also equal to two right angles

So, the angles CBA + ABE = CBA + ABD

Subtract the angle CBA from both sides

Therefore the remaining angle ABE is equal to the remaining angle ABD:

The smaller angle is bigger than the larger angle which is IMPOSSIBLE

So, BE is not in a straining line with CB

Similarly, we can prove that neither is any other straight line except BD

Therefore CB is in a straight line with BD

A

B D

A E

B C D

B

C

B

A

B

Q.E.D

33

Proposition 15

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

EUCLID uses the words vertical angles we would use the words opposite angles

[Translation: If you draw any two straight lines which cut each other the opposite angles,

made by the crossing of the lines, are equal]

Construction

Draw any two lines and let them cross at E

To Prove

AEC = DEB

AED = CEB

Proof

Since AE cuts the line CD making the angles CEA and AED, the angles CEA + AED = 2 right angles

Proposition 13

Again, since the straight line CE cuts the line AB making the angles AED and DEB:

AED + DEB = two right angles (1800) Proposition 13

But CEA + AED also = two right angles

CEA + AED = AED + DEB Postulate 4 and Common Notion 1

Subtract AED from both sides of the equation

CEA = BED Common Notion 3

Similarly it can be proved that CEB = DEA

Q.E.D.

A

D C E

B

34

Proposition 16

In any , if one of the sides be produced, the exterior angle is greater than either of the

interior and opposite angles.

[Translation: If you draw any and extend one of the sides the exterior angle you produce is

bigger than either of the two angles on the far side of the ]

Construction 1

Draw any ABC and extend the side BC to D

To Prove

ACD is greater than either of the interior and opposite angles CBA or BAC

Proof

Bisect AC at E (Proposition 10) and join BE and extend in a straight line to F

Extend BE to F so that EF = BE Proposition 3

Join FC Postulate 1

Extend AC to G Postulate 2

Then, since AE = EC and BE = EF

AE = CE and EB = EF

And AEB = FEC

Because they are vertical angles Proposition 15

Therefore AB = FC

And ABE= CFE

And the remaining angles are equal to the remaining angles respectively, namely those which

the equal sides subtend Proposition 4

Therefore BAE = ECF

A

B C

D

D

E

A

B C

D

D

F

.

35

But ECD > ECF Common Notion 5

Therefore ACD > BAE

Similarly if BC is bisected BCG (equal to ACD - Proposition 15) can be proved greater than

ABC as well

Q.E.D.

36

Proposition 17

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

[Translation: In any , if you add any TWO angles together the sum will always come to less

than two right angles 1800]

Construction

Draw any ABC

To Prove

Any two angles, when added together, must be equal to less than two right angles

Construction

Produce BC to D

Proof

ACD is an exterior angle of the ABC

It is greater than the interior and opposite ABC Proposition 16

Now add the angle ACB to each

angles ACD + ACB are greater than the angles ABC + BCA

But, ACD + ACB is greater than two right angles Proposition 13

ABC + BCA is less than two right angle

Similarly, we can prove that the angles BAC + ACB are less than two right angles and so are the

angles CAB + ABC, as well

A

B C

D

A

B C

D

D

Q.E.D

37

Proposition 18

In any the greater side subtends the greater angle.

[Translation: In any the largest angle is opposite the largest side]

Construction

Draw any ABC with AC bigger than AB

Draw the line BD so that AB = AD Proposition 3

To Prove

ABC is larger than BCA

Proof

ADB i an exterior angle of the BCD

it is greater than the interior and opposite DCB Proposition 16

But ADB = ABD the base angles of an Isosceles are equal

ABD is also greater than the angle ACB

ABC is much greater than the angle ACB

Q.E.D.

A

B C

D

38

Proposition 19

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

[Translation: In any the biggest side is opposite the biggest angle]

Construction

Draw any ABC with the angle ABC greater than the angle BCA

To Prove

The side AC is greater than the side AB

Proof

If AC is not bigger than AB then either:

AC = AB or AC

39

Proposition 20

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

[Translation: In any , if you add the length of any two sides together the length will be bigger

than the length of the third side]

To Prove

In any :

AB + AC > BC

AB + BC > AC

CB + CA > AB

Construction

Extend the line BA to the point D so that DA is equal to CA

Join DC

Proof

Since DA = AC then

ADC = ACD Proposition 5

the BCD > ADC

And, since DCB is a having the angle BCD > angle BDC

And the greater angle is opposite the greater side Proposition 19

DB is greater than BC

But DA is equal to AC (we constructed it that way),

BA + AC > BC

We can prove the other relationships:

AB + BC > CA and

BC + CA > AB

Q.E.D.

A

B C

A

B C

D

40

Proposition 21

If on one of the sides of a , from its extremities, there be constructed two straight line

meeting within the , the straight lines so constructed will be less than the remaining

two sides of the , but will contain a greater angle.

[Translation: If you draw a and choose one of its sides and draw two straight lines to meet at a

point inside the the lengths of the sides drawn added together will add up to less than the

lengths of the two remaining sides, but the angle they make will be bigger than the angle made

by these two sides]

Construction

Draw any ABC

Draw two extra lines within the BD and DC, so that they meet at D within the

To Prove

BD + DC is less than BA + AC, BUT the angle BDC > BAC

Proof

Extend the line BD to E

Since, in any , the lengths of two sides added together are greater than the length of the third

side Proposition 20, then:

In the ABE, the two sides AB+AE > BE

Let EC be added to each

Then BA + AC > BE + EC

Again, since in CED:

CE + ED > CD

Add DB to each side

CE + EB are greater than CD + DB

A

B B B

A

C

D

A

C

D

E

B

41

But BA + AC have been proved to be greater than BE + EC

BA + AC are much greater than BD + DC

Again, since in any triangle the exterior angle is greater than the interior and opposite angle

Proposition 16

, in CDE:

The exterior BDC > CED

For the same reason, in the triangle ABE:

The exterior CEB > BAC

But the angle BDC was proved greater than the CEB

Therefore BDC is much greater than BAC

Q.E.D.

42

Proposition 22

Out of three straight lines, which are equal to three given straight lines, to construct a

: thus it is necessary that two of the straight lines taken together in any manner

should be greater than the remaining one.

[Translation: If you can draw a with three given lengths of line then the lengths of any two

lines added together must be longer than the length of the third line]

Construction

Draw three lines which will be the sides of the remember that two of the straight

lines added together should be longer than the third. Label the lines A, B and C

Draw a straight line DE (it can be of any length)

Mark three points on the line such that DF is the same length as the line A, FG is the

same length as B and GH is the same length as C

With centre F and radius FD draw a circle DKL

With centre G and radius GH draw a circle KLH

Join KF and KG

To Prove

The has been constructed out of the lines A, B and C

A

B

C

F G D

H E

F G D

H E

K

L

43

Proof

The point F is the centre of the circle DKL, so

FD = FK

But FD also equals the line A (we used these line lengths to draw the )

Also, since the point G is the centre of the circle LKH, so

GH = GK

But GH = C (another of the line lengths we used to draw the )

KG is also = 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

Therefore out of the three straight lines KF, FG, GK which are equal to the three given straight

lines A, B, C the KFG has been constructed

Q.E.F

44

Proposition 23

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

given rectilineal angle.

[Translation: So, you are given a straight line and an angle size and you need to construct an

angle of the same size on the line you have been given}

To Start With .....

Draw a straight line AB with the point A marked on it

Draw the angle DCE

To Construct

We have to construct an angle on the line AB, at the point A, an angle equal to the one we have

been given i.e. DCE

Construction

On the straight line CD and CE mark two points D and E anywhere on the two lines

Join DE

Now draw a AFG with the same length sides as those we have just drawn i.e. CD, DE and CE,

so that:

CD = AF, CE = AG and DE = FG Proposition 22

Since the sides DC and CE are equal to the sides FA and AG (we have just drawn them to be so)

and the third side DE = FG (again we have drawn it) then

DCE = FAG Proposition 8

Therefore on the given straight line AB, and at the point A on it, the angle FAG has been

constructed equal to the angle DCE we were given originally Q.E.F

C

D

E

A G B

F

A B

45

Proposition 24

If two s 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.

[Translation: If you have two s which have two sides equal, but one of the included angles is

bigger than the other included angle then the third side will also be larger]

Construction

Draw two s ABC and DEF with two corresponding sides equal i.e. AC = DF and AB = DE

Draw the s such that CAB is bigger than FDE

To Prove

That CB is greater than FE

Construction

Draw the angle EDG equal to the BAC

Proof

The CAB is greater than FDE (we have just drawn it to be so)

Draw the angle EDG so that EDG = BCA Proposition 23

Draw DG equal to either of the lines AC or DF

Draw lines to join EG and FG

We have AB = DE and AC = DG, so BA = ED and AC = DG

A

B

C

D

F

E G

46

And the BAC = EDG

Therefore, the base BC = base EG Proposition 4

Again, since DF = DG the DFG = DFG Proposition 5

Therefore, EFG is much greater than the angle EGF

And, since EFG I a having the angle EFG greater than the EGF and the greater angle is

opposite the greater side Proposition 19:

The side EG is also greater than EF

But, we also have EG = BC

Therefore BC is also greater than EF

Q.E.D.

47

Proposition 25

If two s 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.

[Translation: If you have two s with two sides in one equal to two sides on the other, but the

third side in one is bigger than the third side in the other then the inclusive angle in that one

will be bigger than the inclusive angle in the other]

Construction

Draw two s ABC and DEF, such that

AB = DE and AC = DF

The base EF is greater than the base BC

To Prove

EDF is greater than BAC

Proof

If BAC is NOT greater than EDF then is MUST be either EQUAL to it or LESS than it

The BAC is NOT equal to the EDF, because then the base BC would also have to be equal to

the base EF Proposition 4; but IT IS NOT

Therefore, BCA is NOT EQUAL to EDF

A

B C

D

F

E

48

Neither is BAC less than the EDF, because then the base BC would also have to be less than

the base EF Proposition 24; but IT IS NOT

Therefore, the BAC is NOT less than the angle EDF

But we have also proved that it is NOT EQUAL either

Therefore the BAC is greater than the EDF

Q.E.D.

49

Proposition 26

If two s 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.

[Translation: If two s have two angles in one equal to two angles in the other, and one side

equal to a side in the other (with the side containing the angles or the side opposite one of the

equal angles the remaining sides and angles will be equal} Thi is an examples of proving

congruence between two

Construction

Draw two s ABC and DEF such that:

ABC = DEF and BCA = EFD and

BC = EF

To Prove

AB = DE and AC = DF

And BAC = EDF

Proof

Lets start with proving AB = DE

If AB DE then ONE OF THEM MUST BE BIGGER THAN THE OTHER

Suppose AB is the bigger and let BG be made equal to DE and join GC

Since BG = DE (we have just drawn it so) and BC = EF (we drew the lines equal at the start), then:

GB = DE and BC = EF

A

B C H

G

D

F E

50

And the GBC = DEF

Therefore the base GC = base DF

And the GBC is equal to the DEF, so the remaining angles will be equal to one another

Proposition 4

Therefore, GBC = DEF

But, DFE = BCA (we drew it this way originally)

Therefore BCG = BCA where BCG is bigger than BCA which is clearly nonsense!

Therefore AB is NOT UNEQUAL to DE

And so is EQUAL to it

But BC is also equal to EF, therefore:

AB = DE and BC = EF

And ABC = DEF

Therefore the base AC is equal to the base DF

And the remaining BAC = EDF Proposition 4

Lets now turn our attention to the sides AB and DE

If we know that AB = DE then we need to prove that AC = DF and BC = EF and theta the

remaining BAC = EDF

Proof

Suppose BC EF then one of them MUST be larger than the other

Suppose BC is the larger and mark off BH equal to EF and draw the line AH

Now, since BH = EF (we have just drawn it so) and AB = DE (part of the original construction)

then we have:

AB = DE and BH = EF and they contain equal angles

Therefore the line AH = DF

And the ABH the DEF

So, the remaining angles in the one will be equal to the remaining angles in the other

Proposition 4

51

Therefore, BHA = EFD

But the EFD = BCA

Therefore, in the AHC the exterior angle BHA is equal to the interior and opposite angle BCA

which is IMPOSSIBLE! Proposition 16

Therefore BC EF

And so MUST be equal to it

But, AB = DE

Therefore AB = DE and BC = EF and they contain equal angles

Therefore AC = DF

The ABC DEF

And the remaining BAC = EDF Proposition 4

Q.E.D

52

Proposition 27

If a straight line falling on two straight lines makes the alternate angles equal to one

another, the straight lines will be parallel to one another

[Translation: If you have two straight lines and a line cutting them, and the alternate angles

made by the cutting line, are equal the two lines are parallel]

Construction

Draw two straight lines AB and CD

Draw the line EF to cut the two straight lines at E and F

Suppose we measure the alternate angles AEF and EFD and find they are the same

To Prove

AB is parallel to CD

Construction

Suppose that AB is not parallel to CD

Then we can extend AB and CD and they will meet at, say, G

Proof

If AB is NOT parallel to CD then the two lines must meet either in the direction of BD or in the

direction of AC

Lets assume they will meet in the direction of BD at a point G and draw the appropriate lines

BG and DG

In the GEF the exterior angle AEF is equal to the interior and opposite angle EFG, which is

impossible

Proposition 16

Therefore AB and CD when produced will NOT meet in the direction of BD

Similarly, we can prove that they will NOT meet towards AC

But, straight lines which do not meet in either direction are PARALLEL

Therefore AB is parallel to CD

A B

C D

G

E

F

Q.E.D.

53

Proposition 28

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 of the same side

equal to two right angles, the straight lines will be parallel to one another.

[Translation: If you draw two straight lines and then cut them with a third and the exterior

angle is equal to the interior and opposite angle on the same side, or the angles on the same

side equal two right angles (1800) then the straight lines are parallel to each other]

Construction

Draw two lines AB and CD

Draw a line EF to cut AB and CD at G and H

Let the straight line EF which cuts the line AB and CD make exterior angle EGB = GHD

Or BGH + GHD = 1800

To Prove

AB parallel CD

Proof

EGB = GHD (weve drawn it this way) and

EGB = AGH Proposition 15

AGH = GHD and they are alternate

Therefore AB is parallel to CD Proposition 27

A B

C D

E

G

H

F

54

Similarly, since

BGH + GHD = 1800 and

AGH + BGH = 1800

The angle AGH + BGH = BGH + GHD

Subtract the angle BGH from both sides:, to get

AGH = GHD

And they are alternate angles

Therefore AB is parallel to CD Proposition 27

Q.E.D.

55

Proposition 29

A straight line falling on parallel 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.

[Translation: If you draw two parallel straight lines and cut them both with a single straight line

the exterior angles will be equal to the opposite interior angles and the interior angles on the

same side will be equal to two right angles (1800)]

Construction

Draw two straight parallel lines AB and CD

Draw a straight line EF which cuts both lines

To prove

AGH = GHD

EGB = GHD

BGH + GHD = 1800

Proof

Suppose AGH GHD then one of them is bigger than the other

Suppose AGH is bigger than GHD

If we add the BGH to both AGH and GHD

Then AGH + BGH is bigger than BGH + GHD

A B

C D

E

G

H

F

A B

C D

E

G

H

F

56

But, AGH + BGH = 1800 Proposition 13

Therefore BGH + GHD < 1800

But, straight lines produced indefinitely from angles less than 1800 MEET Postulate 5

Therefore the lines AB and CD, if produced indefinitely, will meet

But THEY WILL NOT meet because we have drawn them not to meet!

Therefore, AGH is NOT equal to GHD

THEREFORE IT IS EQUAL TO IT

Again, the AGH = EGB Proposition 15

Therefore the EGB = GHD Common Notion 1

Suppose we add the angle BGH to both EGB and GHD

Then EGB + BGH = BGH + GHD Common Notion 2

But the EGB + BGH = 1800 Proposition 13

Therefore BGH + GHD also = 1800

Q.E.D.

57

Proposition 30

Straight lines parallel to the same straight line are also parallel to one

another.

[Translation: If straight lines are all parallel to another straight line they are also all parallel to

each other]

Construction

Draw a straight line EF

Draw two further straight lines AB and CD which are parallel to the line EF

To prove

AB is parallel to CD

Proof

Draw a line GK which cuts the three lines

Since the line GK has cut the parallel lines AB and EF

The AGK = GHF Proposition 29

Also, since the straight line GK cuts the straight lines EF and CD:

The GHF = GKD Proposition 29

But the angle AGK has already been proved to be equal to GHF

Therefore AGK = GKD Common Notion 1

And they are alternate

Therefore AB is parallel to CD

Q.E.D.

E F

A B

C D

E F

G

H

K

58

Proposition 31

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

[Translation: you are given a point and a straight line (not through the point) draw a line thorugh the

point which is parallel to the line given]

Construction

Mark any point and label it A

Draw any straight line and label it BC

To Construct

A straight line through the point A parallel to the straight line BC

Construction

Make a point D anywhere on BC

Join the points D and A

On the straight line DA and at the point A on it construct the angle DAE the same size as the

angle ADC Proposition 23

Extend the straight line EA to F

Now, since the straight line AD cuts the two straight lines BC and EF and makes the alternate

angles EAD and ADC equal to one another:

Therefore the line EAF is parallel to the line BC Proposition 27

Therefore we have drawn, through the point A, a line (EAF) parallel to the given straight line BC

Q.E.D.

.A

B C

A

B C D

F E

59

Proposition 32

In any , 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 are equal to two right

angles.

[Translation: In a ABC, extend BC to D, then f = g + h and g + h + k = 1800]

Construction

Draw any ABC and extend BC to D

To Prove:

ACD = ABC + BAC

And

ABC + BCA + CAB = two right angles (1800)

Proof

Draw a straight line from C parallel to the straight line AB Proposition 31

Then, since AB is parallel to CE and the line AC cuts both lines - BAC = ACE Proposition 29

Also, since AB is parallel CE and the straight line BD cuts them both:

The exterior ECD = ABC Proposition 29

But we have already proved that ACE = ACE

Therefore the whole ACD = BAC + ABC

Now add the ACB to each side

We get:

B C D

A

B C D

A E

f

g

h k

60

ACD + ACB = ABC + BCA +CAB

But we know that ACD + ACB = 1800 Proposition 13

Therefore ABC + BCA +CAB also = 1800

Q.E.D.

61

Proposition 33

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.

[Translation: If any two equal parallel straight lines BA and CD are joined at their respective ends, the

lines joining are also equal and parallel]

Construction

Draw any two equal straight lines (AB and CD) parallel to each other

Join the ends of the lines BD and AC

To Prove

Lines BD and AC are equal and parallel

Proof

Draw in the line BC

We know that AB is parallel to CD and that BC cuts both lines, so

The alternate angles ABC = BCD Proposition 29

And, since AB = CD (this is what we drew in the first place) and BC is common to the two s ABC

and BDC

AB = DC and BC = CB and ABC = BCD

Therefore the line AC = BD

And the ABC DCB

And so the remaining angles will be equal to the remaining angles in the two s Proposition

4

Therefore ACB = CBD

And, since BC cuts the two lines BA and DC and makes the alternate angles equal to one

another:

AC is parallel to BD Proposition 27

And it was also proved equal to it

B A

D C

B A

D C

Q.E.D.

62

Proposition 34

In parallelogrammic areas the opposite sides and angles are equal to one another, and

the diameter bisects the areas.

[Translation: In any parallelogram ABCD the sides AB and CD are equal and so are the sides AC and BD. In

addition A = D and C = B. Finally, either diagonal bisects the area of the parallelogram e.g.

CB)

Construction

Draw any parallelogram ACDB and draw either of the diagonals BC or AD (or diameters

as Euclid calls them)

To Prove

Opposite angles of the parallelogram are equal (ACD = DBA and CAB = CDB)

The diagonal BC bisects the parallelogram

Proof

Since the lines AB and CD are parallel and the line BC cuts them:

The alternate angles ABC = BCD Proposition 29

Also, since AC is parallel to BD and BC cuts both lines

The alternate angles ACB = CBD Proposition 29

Therefore ABC and DCB are two s having two pairs of angles equal (ABC = DCB and BCA =

CBD) and a common side equal BC

Therefore they will have the remaining sides equal to the remaining sides respectively and the

remaining angles equal to the remaining angles respectively Proposition 26

There fore AB = CD and AC = BD

And, BAC = CDB

And, since ABC = BCD

And the CBD = ACB

A B

C D

63

The whole angle ABD = whole ACD Common Notion 2

And the angle BAC was also proved = CDB

Therefore in the parallelogram the opposite sides and angles are equal to one another

Now we need to prove that the diagonal bisects the parallelogram

We have AB = CD

And BC is common

The sides AB = DC and BC = CB respectively

And the angle ABC = BCD

Therefore AC = DB

And the s ABC DCB Proposition 4

Therefore the diagonal BC bisects the parallelogram ACBD

Q.E.D.

64

Proposition 35

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

another

[Translation: If you draw two parallelograms BCDA and BCFE so that they are on the same base (BC) and

between the same parallel lines then they have equal areas]

Construction

Draw two parallelograms ABCD and EBCF on the same base (BC) and within the same

parallel lines AF and BC

To Prove

ABCD EBCF

Proof

Since ABCD is a parallelogram:

AD = BC Proposition 34

For a similar reason:

EF = BC

So that AD is also equal to EF Common Notion 1

And, since DE is common, the whole of AE = the whole of DF Common Notion 2

But AB = DC Proposition 34

Therefore EA = FD and AB = DC

And FDC = EAB, the exterior to the interior Proposition 29

Therefore EB = FC and the EAB FDC Proposition 4

Lets subtract the DGE from both s

Then the trapezium ABGD which is left is equal to the trapezium EGCF Common Notion 3

Lets add the GBC to each of the trapezia ABGD and EGCF

Then, the whole parallelogram ABCD = the whole parallelogram EBCF

A D

B C

E F

G G

A D

B C

E F

Q.E.D.

65

Proposition 36

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

another

[Translation: If you draw two parallelograms BCDA and FGHE with the same size base (i.e. BC = FG) (but

NOT necessarily the same base) and between the same parallels then their areas are equal]

Construction

Draw two parallel line AH and BG

Draw any two parallelograms ABCD and EFGH with the same size bases (BC = FG) within

the parallels.

To Prove

Parallelogram ABCD parallelogram EFGH

Proof

Draw the lines BE and CH

Then, since BC = FG (we drew the parallelograms on equal bases)

And FG = EH (opposite sides of a parallelogram)

BC is also = EH Common Notion 1

But they are also parallel and EB and HC join them

But straight lines joining (at the ends of the lines) equal and parallel straight lines are equal and

parallel Proposition 33

Therefore EBCH is a parallelogram Proposition 34

And is equal to ABCD because it has the same base BC and is in the same parallels BC and AH

Proposition 35

For the same reason EFGH EBCH Proposition 35

Therefore the parallelogram ABCD parallelogram EFGH

A

B C

D E H

F G

A

B C

D E H

F G

Q.E.D.

66

Proposition 37

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

[Translation: If you draw two s ABC and DBC, so that they have the same base (BC) and are drawn

between the same parallels then they have the same area]

Construction

Draw two parallel lines Ad and BC

Draw two s ABC and DBC so that they have the same base (BC) and are within the

parallel lines AD and BC

To Prove

ABC DBC

Proof

Extend the line ED in both directions to E and F

Draw the line BE so that it is parallel to CA

Draw the line CF so that it is parallel to BD

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

on the same base BC and in the same parallels BC and EF Proposition 35

Also, the ABC is half of the parallelogram EBCA because the diagonal bisects it Proposition

34

And the DBC is half of the parallelogram DBCF because the diagonal bisects it Proposition

34

But the halves of equal things are equal to one another

Therefore the ABC DBC

Q.E.D.

A

B C

D A

B C

D E F

67

Proposition 38

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

[Translation: This proposition is similar to Prop 37 except in this case we can have TWO

DIFFERENT bases BC and EF, so long as they are the SAME SIZE and the s will be equal]

Construction

Draw two s ABC and DEF with the same size bases (BC and EF) between two parallel

lines

To Prove

ABC = DEF

Proof

Extend AD in both directions to G and H

Draw a line BG so that it is parallel to CA Proposition 31

Draw another line FH so that it is parallel to DE

Then we have the two shapes GBCA and DEFH are parallelograms, which gives us:

GBCA DEFH because they are on equal bases BC and EF and between the same parallels BF

and GH Proposition 36

Also the ABC is half of the parallelogram GBCA because the diagonal bisects it Proposition 34

And the FED is half the parallelogram DEFH because the diagonal bisects it Proposition 34

But the halves of equal things are equal to one another

Therefore the ABC DEF

Q.E.D.

A

B C

D

E F

A

B C

D

E F

G H

68

Proposition 39

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

parallels.

[Translation: If you draw two s ABC and DBC which are equal in area, and have the same base (BC)

then they are drawn between the same parallel lines]

Construction

Draw two s ABC and BDC so that:

They have the same base (BC)

They are equal in area

To Prove

The two s ABC and BDC are located between parallel lines

Proof

Draw the line AD

AD is parallel to BC because, if it isnt draw the line AE from A parallel to the straight line BC

Proposition 31 and draw the line EC

Then the ABC is equal to the EBC because it is on the same base BC and within the same

parallels Proposition 37

But ABC = DBC (we drew them as equal s)

Therefore DBC is also equal to EBC Common Notion 1

This implies that the greater is equal to the les which is impossible!

Therefore AE is NOT parallel to BC

Similarly we can prove that neither is any other straight line except AD

Therefore AD is parallel to BC

Q.E.D.

A

C B

D A

C B

D

E

69

Proposition 40

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

[Translation: If you draw two s ABC and DCE so they have the same area and on equal bases (BC = CE)

then they are drawn between the same parallel lines]

Construction

Draw two s which are:

Equal in area

Have the same sized base

Label these s ABC and CDE

To Prove

They sit between two parallel lines

Proof

Join AD

We will first prove that AD is parallel to BE

Suppose it isnt

Draw the straight line AF from A parallel to BE (Proposition 31) and also join FE

Therefore the ABC = FCE because they are on equal bases BC and BE and between the same

parallels BE and AF Proposition 38

But, the ABC = DCE (we drew them to be equal)

Therefore the DCE is also equal to the FCE Common Notion 1

Which gives us the contradiction the greater is equal to the less which is impossible

Therefore AF is NOT parallel to BE

Similarly, we can prove that neither is any other straight line except AD

Therefore AD is parallel to BE

Q.E.D.

A

C B

D

E

A

C B

D

E

F

70

Proposition 41

If a parallelogram have the same base with a and be in the same parallels, the

parallelogram is double of the .

[Translation: If you draw a parallelogram ABCD and a EBC so that they have the same base

(BC) and are drawn between the same parallel lines then the area of the parallelogram is twice

the area of the ]

Construction

Draw two parallel lines AE and BC

Draw a parallelogram ABCD between the two parallels

Draw a BCE so that it has the same base (BC) as the parallelogram

To Prove

The parallelogram is twice the area of the BEC

Proof

Draw a the diagonal line AC

Then the ABC EBC because it is on the same base BC and in the same parallels BC and AE

Proposition 37

But the parallelogram ABCD is double the ABC because the diagonal AC bisects it

Proposition 34

So, the area of the parallelogram is also double the area of the EBC

Q.E.D

A

C B

D E A

C B

D E

71

Proposition 42

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

[Translation: You are given any angle. You are required to construct, on that angle, a parallelogram

equal in area to a given triangle

Construction

Draw any ABC

Draw any angle D

To Construct

On the given angle D a parallelogram equal in area to the given

The Construction

Bisect BC at E and join AE

On the straight line EC, at the point E, construct the CEF equal to the D Proposition 23

Draw a line AG from A parallel to EC, and Proposition 31

Draw a line CG from C parallel to EF

Then FECG is a parallelogram

And, since BE = EC:

ABE = in area AEC because they are on equal bases BE and EC and between the same

parallels BC and AG Proposition 38

Therefore the ABC is double the AEC

But the parallelogram FECG is also double the AEC, because it has the same base as it and is

within the same parallels

Therefore the parallelogram FECG = ABC

And it has the CEF equal to the D

Therefore the parallelogram FECG has been constructed equal to the given ABC and the CEF

which is equal to D

A

C B

D

A

C B E

F G >

>

>

>

Q.E.F.

72

Proposition 43

In any parallelogram the complements of the parallelogram about the diameter are

equal to one another.

[Translation: If you construct any parallelogram, and draw one of the diameters, you will be able to

draw an infinite number of parallelograms which are similar to the original (FKGC in the diagram) and

the spaces left (the complements as Euclid calls them) will be equal]

Construction

Draw a parallelogram ADCB and its diagonal AC (or diameter as Euclid calls it)

Using the line AC as the diagonal draw any parallelogram AHKE

Again using line AC as the diagonal draw another parallelogram KFCG to fill the

remainder of the big parallelogram

To Prove

The area of the parallelogram HDFK is equal to the area of the parallelogram EKGB (the

complements, as Euclid calls them)

Proof

Since ABCD is a parallelogram and AC is a diagonal the ABC = ACD Proposition 34

Also, since AHKE is a parallelogram and AK a diagonal the AEK = AHK

For the same reason:

The KFC = KGC

Since AEK = AHK, and

KFC = KGC

The AEK + KGC = AHK + KFC Common Notion 2

And the whole ABC is also equal to the whole ADC

Therefore the complement EKGB = complement