Euclid.

# A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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[Axiom 12 has been objected to on the double ground that it cannot
be considered self-evident, and that its truth may be deduced from
simpler principles. It is employed for the first time in the 29th Pro-
position of Book L, where a short discussion of the difficulty will be
found.

The converse of this Axiom is proved in Book I. Proposition 17.]

INTBODUCTOKY,

INTRODUCTOKY.

Plane Geometry deals with the properties of all lines and
figures that may be drawn upon a plane surface.

Euclid in his first Six Books confines himself to the properties
of straight lines, rectilineal figures, and circles.

The Definitions indicate the subject-matter of these books:
the Postulates and Axioms lay down the fundamental principles
which regulate all investigation and argument relating to this
subject-matter.

Euclid's method of exposition divides the subject into a
number of sej)arate discussions, called propositions; each pro-
position, though in one sense complete in itself, is derived from
results previously obtained, and itself leads up to subsequent
propositions.

Propositions are of two kinds. Problems and Theorems.

A Problem i3roposes to effect some geometrical construction,
such as to draw some particular line, or to construct some re-
quired figure.

A Theorem proposes to demonstrate some geometrical truth.

A Proposition consists of the following parts :

The General Enunciation, the Particular Enunciation, the
Construction, and the Demonstration or Proof.

(i) The General Enunciation is a preliminary statement,
describing in general terms the purpose of the proposition.

In a problem the Enunciation states the construction which
it is proposed to effect: it therefore names first the Data, or
things given, secondly the Qusesita, or things required.

In a theorem the Enunciation states the property which it
is j)roposed to demonstrate : it names first, the Hypothesis, or
the conditions assumed ; secondly, the Conclusion, or the asser-
tion to be X)rovcd.

10 Euclid's elements.

(ii) The Paaijicular Enunciation repeats in special terms

(iii) The Constmction then directs the drawing of such
straight lines and circles as may be required to effect the purpose
of a problem, or to prove the truth of a theorem.

(iv) Lastly, the Demonstration proves that the object pro-
posed in a problem has been accomplished, or that the propei-ty
stated in a theorem is true.

Euclid's reasoning is said to be Deductive, because by a con-
nected chain of argument it deduces new truths from truths

The initial letters q.e.f., placed at the end of a problem,
stand for Quod erat Faciendum, which was to he done.

The letters q. e. d. are appended to a theorem, and stand for
Quod erat Demonstrandum, which was to be proved.

A Corollary is a statement the truth of which follows readily
from an established proposition ; it is therefore appended to the
proposition as an inference or deduction, which usually requires
no further proof.

The following symbols and abbreviations may be employed
in writing out the propositions of Book L, though their use is not
recommended to beginners.

.-. for

therefore.

par^ (or

11) fo^

â€˘ parallel,

~ J)

is, or are, equal

to,

par"

Â»

parallelogram.

^

angle,

sq.

?)

square,

rt. L â€ž

right angle,

rectil.

Â»

rectilineal.

A â€ž

triangle,

st. line

Â»j

straight line,

perp. â€ž

perpendicular.

pt.

5)

point;

and all obvious contractions of words, such as opp., adj., diag.
&c., for opposite, adjacent, diagonal, &c.

BOOK I. PROP. 1. 11

SECTION I.

Proposition 1. Problem.

To describe an equilateral triangle 07h a given Jlnite
straight line.

Let AB be the given straight line.
It is required to describe an equilateral triangle on AB.
Construction. From centre A, with radius AB, describe
the circle BCD. Post. 3,

From centre B, with radius BA, describe the circle ACE.

Post. 3.

From the point C at which the circles cut one another,

draw tlie straight lines CA and CB to the points A and B.

Post. 1.
Then shall ABC be an equilateral triangle.

Proof. Because A is the centre of the circle BCD,

therefore AC is equal to AB. Def. 11.

And because B is the centre of the circle ACE,

therefore BC is equal to BA. Def. 1 1 .

But it has been shewn that AC is equal to AB ;
therefore AC and BC are each equal to AB.
But things which are equal to the same thing are equal
to one another. Ax. 1.

Therefore AC is equal to BC.
Therefore CA, AB, BC are equal to one another.
Therefore the triangle ABC is equilateral ;
and it is described on the given straight line AB. q.e. F.

12 euclid's elements.

Proposition 2.. Problem.

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

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

Construction. JoinAB; Post. 1.

and on AB describe an equilateral triangle DAB. i. 1.
From centre B, with radius BC, describe the circle CGH.

Post. 3.
Produce DB to meet the circle CGH at G. Post. 2.
From centre D, with radius DG, describe the circle GKF.
Produce DA to meet the circle GKF at F. Post. 2.
Then AF shall be equal to BC.

Proof. Because B is the centre of the circle CGH,

therefore BC is equal to BG. Def. 11.

And because D is the centre of the circle GKF,

therefore DF is equal to DG ; Dcf. 11.

and DA, DB, parts of them are equal; Def. 19,

therefore the remainder AF is equal to the remainder BG.

Ax. 3.
And it has been shewn that BC is equal to BG ;
therefore AF and BC are each equal to BG.
But things which are equal to the same thing are equal
to one another. Ax. 1.

Therefore AF is equal to BC ;
and it has been drawn from the given point A. q. e. f.

[This Proposition is rendered necessary by the restriction, tacitly
imposed by Euclid, that compasses shall not be used to transfer
distances.]

BOOK I. PRO". 3. 13

Proposition 3. Problem.

From the greater of tivo given straight li7ies to cut off a
fart equal to the less.

Let AB and C ])e the two given straight lines, of wliich
AB is tlie greater.

It is required to cut off from AB a part equal to C.

Construction. From the point A draw the straight line

AD equal to C ; i. 2.

and from centre A, wit a radius AD, describe the circle DEF,

meeting AB at E. Post. 3.

Then AE shall be equal to C.

Proof, Because A is the centre of the circle DEF,

therefore AE is equal to AD. Def 11.

But C is equal to AD. Constr.

Therefore AE and C are each equal to AD.
Therefore AE is equal to C ;
and it has been cut off from the given straight line AB.

Q.E.P

EXERCISES.

1. On a given straight line describe an isosceles triangle having
each of the equal sides equal to a given straight line.

2. On a given base describe an isosceles triangle having each of
the equal sides double of the base.

3. In the figure of i. 2, if AB is equal to BC, shew that D, the
vertex of the equilateral triangle, ^Yill fall on the circumference of the
circle CGH.

14 EUCLID'S ELEMENTS.

Ohs. Every triangle has six parts, namely its three sides
and three angles.

Two triangles are said to be equal in all respects, when
they can be made to coincide with one another by superposition
(see note on Axiom 8), and in this case each part of the one ia
equal to a corresponding part of the other.

Proposition 4. Theorem.

If two triangles liave ttoo sides of the one equal to two
sides of the other, ea^h to each, and have also tlie angles
contained hy those sides eqiial; then shall their bases or third
sides he equal, and the triangles slmll he equal in area, and
their remaining angles shall he equal, each to each, namely
tJiose to which the equal sides are opposite : timt is to say, the
triangles shall he equal in all respects.

D

_ E
C

Let ABC, DEF be two triangles, which have the .side AB

equal to the side DE, the side AC equal to the side DF, and

the contained angle BAC equal to the contained angle EDF.

Then shall the base BC be equal to the base EF, and the

triangle ABC shall be equal to the triangle DEF in area;

and the remaining angles shall be equal, each to each, to

which the equal sides are opposite,

namely the angle ABC to the angle DEF,

and the angle ACB to the angle DFE.

For if the triangle ABC be applied to the triangle DEF,

so that the point A may be on the point D^

and the straight line AB along the straight line DE,

then because AB is equal to DE, Hyp.

therefore the point B must coincide with the point E.

BOOK I. PROP. 4. 15

And because AB falls along DE,
and the angle BAG is equal to the angle EDF, Hyp.
therefore AC must fall along DF.
And because AC is equal to DF, Hyp.

therefore the point C must coincide with tlie point F.
Then B coinciding with E, and C with F,
the base BC must coincide with the base EF;
for if not, two straight lines would enclose a space ; which
is impossible.- Ax. 10.

Thus the base BC coincides with the base EF, and is
therefore equal to it. Ax. %.

And the triangle ABC coincides with tlie triangle DEF,
and is therefore equal to it in area. Ax. 8.

And the remaining angles of the one coincide with the re-
maining angles of the other, and are therefore equal to them,
namely, the angle ABC to the angle DEF,
and the angle ACB to the angle DFE.
That is, the triangles are equal in all respects. q. e. d.

Note, It follows that two triangles which are equal in their
several parts are equal also in area; but it should be observed that
equality of area in two triangles does not necessarily imply equality in
their several parts: that is to say, triangles may be equal in area,
without being of the same

Two triangles which are equal in all respects have identity of form
and magnitude, and are therefore said to be identically equal, or
congruent.

The following application of Proposition 4 anticipates
tlie chief difficulty of Proposition 5.

In the equal sides AB, AC of an isosceles triangle
ABC, the points X and Y are taken, so that AX
is equal to AY ; and BY and CX are joined,
^hew that BY is equal to CX.

In the two triangles XAC, YAB,
XA is equal to YA, and AC is equal to AB ; Hyp.
that is, the two sides XA, AC are equal to the two

sides YA, AB, each to each;
and the_ angle at A, which is contained by these
sides, is common to both triangles :
therefore the triangles are equal in all respects ;
so that XC is equal to YB.

16 euclid's elements.

Proposition 5. Theorem.

lite angles at the base of an isosceles triangle are equal
to one another; and if the equal sides he produced^ the
angles on the other side of the hose shall also he equal to one
anotJier.

Let ABC be an isosceles triangle, having the side AB
equal to the side AC, and let the straight lines AB, AC be
produced to D and E :

then shall the angle ABC be equal to the angle ACB,
and the angle CBD to the angle BCE.

Construction. In BD take any point F;
and from AE the greater cut off AG equal to AF the less. i. 3.^
Join FC, GB.

Proof. Then in the triangles FAC, GAB,

( FA is equal to GA, Constr.

Because I and AC is equal to A B, /%?.

jalso the contained angle at A is common to the
I two triangles ;
therefore the triangle FAC is equal to the triangle GAB in
all respects ; i. 4.

that is, the base FC is equal to the base GB,
and the angle ACF is equal to the angle ABG,
also the angle AFC is equal to the angle AGB.
Again, because the whole AF is equal to the whole AG,
of which the parts AB, AC are equal, ^J^yp-

therefore the remainder BF is equal to the I'einaindei* CG.

BOOK I. PROP. 5. 17

Then in the two triangles BFC, CGB,

r BF is equal to CG, Proved.

,, and PC is equal to GB, Proved.

I also the contained angle BFC is equal to the

\ contained angle CGB, Proved.

therefore the triangles BFC, CGB are equal in all respects;

so that the angle FBC is equal to the angle GCB,

and the angle BCF to the angle CBG. i. 4.

Now it has been shewn that the whole angle ABG is equal

to the whole angle ACF,
and that parts of these, namely the angles CBG, BCF, are

also equal ;
therefore the remaining angle ABC is equal to the remain-
ing angle ACB ;

and these are the angles at the base of the triangle ABC.
Also it has been shewn that the angle FBC is equal to the

angle GCB ;
and these are the angles on the other side of the base, q.e.d.

Corollary. Hence if a triangle is equilateral it is
also equiangular.

Definition. Each of two Theorems is said to be the Con-
verse of the other, when the hypothesis of each is the conclusion
of the other.

It will be seen, on comparing the hypotheses and conclusions of
Props, o and 6, that each proposition is the converse of the other.

Note. Proposition 6 furnishes the first instance of an indirect
method of "proof, frequently used by Euclid. It consists in shewin*;
that an absurdity must result from supposing the theorem to be
otherwise than true. This form of demonstration is known as the
Reductio ad Absurdum, and is most commonly employed in establish-
ing the converse of some foregoing theorem.

It must not be supposed that the converse of a true theorem is
itself necessarily true : for instance, it will be seen from Prop. 8, Cor.
that if two triangles have their sides equal, each to each, then their
angles will also be equal, each to each ; but it may easily be shewn by
means of a figure that the converse of this theorem is not necessarily
true.

H. E.

18 EUCLID'S ELEMENTS.

Proposition 6. Theorem.

If two angles of a triant/le he equal to otie another^ then
the sides also which subtend, or are opposite to, the equal
angles, shall he equal to one another.

Let ABC be a triangle, having the angle ABC equal to
the angle ACB :

then shall the side AC be equal to the si^e AB.

Construction. For if AC be not equal to AB,
one of them must be greater than the other.
If possible, let AB be the greater;
and from it cut off BD equal to AC. i. 3.

Join DC.

Proof. Then in the triangles DBC, ACB,

[ DB is equal to AC, Conslr.

and BC is common to both,

also the contained angle DBC is equal to the

I contained angle ACB; iiyj>-

therefore the triangle DBC is equal in area to the triangle

ACB, 1. 4.

the part equal to the whole ; which is absurd. Ax. 9.

Therefore AB is not unequal to AC ;

that is, AB is equal to AC. q.kd.

Corollary. Hence if a triangle is equiangular it is
also equilateral.

Because

BOOK I. nior. 7. 19

Proposition 7. Theorem.

Oyi the same base, and on the same side of it, there
cannot he tivo triangles havirig their sides which are termi-
nated at one extremity of the base equal to one another, and
lihevnse those which are terminated at the other extremity
equal to one another.

If it be possible, on the same base AB, and ou the same
side of it, let there Ije two triangles ACB, ADB, having their
sides AC, AD, which are terminated at A, equal to one
another, and likewise their sides BC, BD, which are termi-
nated at B, equal to one another.

Case I. When the vertex of each triangle is without
the other triangle.

Construction. Join CD. Post. 1.

Proof Then in the triangle ACD,

because AC is equal to AD, U^VP-

therefore tlie angle ACD is equal to the angle ADC. I. 5.

But the whole angle ACD is greater than its part, the

angle BCD,
therefore also the angle ADC is greater than the angle BCD ;
still more then is the angle BDC greater than the angle
BCD.

Again, in the triangle BCD,
because BC is equal to BD, ^fyp-

therefore the angle BDC is equal to the angle BCD: i. 5
but it was shewn to be greater ; which is impossible.

2â€”2

20 EUCLID'S ELEMENTS.

Cask II. Wlieii one of the vertices, as D, is within
tlio other triangle ACB. ^

Construction. As before, join CD ; Post. 1.

and produce AC, AD to E and F. Post. 2.

Tlien in the triangle ACD, because AC is equal to AD, Hyp.

therefore the angles ECD, FDC, on the other side of the

base, are equal to one another. I. 5.

But the angle ECD is greater than its part, the angle BCD;

therefore the angle FDC is also greater than the angle

BCD:
still more then is the angle BDC greater than the angle
BCD.

Again, in the triangle BCD,
because BC is equal to BD, Hy^j.

therefore the angle BDC is equal to the angle BCD : I. 5.
but it has been shewn to be greater ; which is impossible.

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

Therefore AC cannot be equal to AD, and at the same
time J BC equal to BD. q.e.d.

Note. The sides AC, AD are called conterminous sides ; similarly
the sides BC, BD are conterminoas.

Proposition 8. Theorem.

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

9
BOOK I. PROP. 8. 21

Let ABC, DEF be two triangles, having tlie two sides
BA, AC equal to the two sides ED, DF, each to each, namely
BA to ED, and AC to DF, and also the base BC equal to the
base EF: ' '

then shall the angle BAC be equal to the angle EDF.

Proof. For if the triangle ABC be applied to the
triangle DEF, so that the point B may be on E, and tlie
.straight line BC along EF ;

then because BC is equal to EF, ^yP'

therefore the point C must coincide with the point F.

Then, BC coinciding with EF,

it follows that BA and AC must coincide with ED and DF :

for if not, they would have ^ different situation, as EG, G F :

then, on the same base and on the same. side of it there

Â» would be two triangles having their conterminous sides
equal.
But this is impossible. i. 7.

Therefore tlie sides BA, AC coincide with the sides ED, DF.
That is, the angle BAC coincides with the angle EDF, and is
therefore equal to it. Ax. 8.

PQ.E.D.
Note. In this Proposition the three sides of one triangle are
given equal respectively to the three sides of the other; and from
this it is shewn that the two triangles may be made to coincide with
one another.

Hence we are led to the following important Corollary.

Corollary. If in two triangles the three sides of the
one are equal to the three sides of the other, each to each,
then the triangles are equal in all respects.

22

KUCLID'S KLEMENTS.

The following proof of Prop. 8 is worthy of attention as it is inde-
pendent of Prop. 7, which frequently presents difficulty to a beginner.

Proposition 8. Alternative Proof.
A D

Let ABC and DEF be two triangles, which have the sides BA, AC
equal respectively to the sides ED, DF, and the base BC equal to the
base EF :

then shall the angle BAG be equal to the angle EDF.
For apply the triangle ABC to the triangle DEF, so that B may
fall on E, and BC along EF, and so that the point A may be on the
side of EF remote from D,

then C must fall on F, since BC is equal to EF. â€˘
Let A'EF be the new position of the triangle ABC.

If neither DF, FA' nor DE, EA' are in one straight line,
join DA'.

Case I. When DA' intersects EF.

Then because ED is equal to EA',
therefore the angle EDA' is equal to the angle EA'D. i, r>.

Again because FD is equal to FA',
therefore the angle FDA' is equal to the angle FA'D. i. 5.
Hence the whole angle EDF is equal to the whole angle EA'F ;
that is, the angle EDF is equal to the angle BAC.

Two cases remain which may be dealt with in a similar manner:
namely.

Case II. When DA' meets EF produced.

Case III. When one pair of sides, as DF, FA', are in one straight
line.

book i. prop. 9. 23

Proposition 9. Problem.

To bisect a given angle, that is, to divide it into two equal
parts.

A

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

Construction. In AB take any point D;

and from AC cut off AE equal to AD. i. 3.

Join DE;
and on DE, on the side remote from A, describe an equi-
lateral triangle DEF. i. 1.
Join AF.
Then shall the straight line AF bisect the angle BAG.

Proof. For in the two triangles DAF, EAF,

( DA is equal to EA, Constr.

p I and AF is common to both;

Jjecause \^ ^^^ ^j^.^^ ^-^^ ^^P -^ ^^^^^^ ^^ ^j^^ ^j^.^,^ ^.^^^

I EF; Def. 19.

therefore the angle DAF is equal to the angle EAF. i. 8.
Therefore the given angle BAG is bisected by the straight
line AF. Q.e.f.

EXERCISES.

^1. If in the above figure the equilateral triangle DFE were de-
scribed on the same side of DE as A, what different cases would arise?
And under what circumstances would the construction fail?

2. In the same figure, shew that AF also bisects the angle DFE.

3. Divide an angle into four equal parts.

24 Euclid's elements.

Proposition 10. Problem.

To bisect a given finite straight linej that is, to divide it
into two equal 2Jarts.

Let AB be the given straight line :
it is required to divide it into two equal parts.

Constr. On AB describe an equilateral triangle ABC, i. 1.
and bisect the angle ACB by the straight line CD, meeting
AB at D. I. 9.

Tlien shall AB be bisected at the point D.

P7'oo/. For in the triangles ACD, BCD,

r AC is equal to BC, Dqf. 19.

-r, and CD is common to both :

I also the contained angle ACD is equal to the con-
[ tained angle BCD; Constr.

Therefore the triangles are equal in all respects:
so that the base AD is equal to the base BD. i. 4.
Therefore the straight line AB is bisected at the point D.

Q. E, F.

EXERCISES.

1. Shew that the straight line which bisects the vertical angle of
an isosceles triangle, also bisects the hase.

2. On a given base describe an isosceles triangle such that the
sam of its equal sides may be equal to a given straight line.

book i. prop. 11. 25

Proposition 11. Problem.

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

Because

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

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

CoTistruction. In AC take any point D,

and from CB cut off CE equal to CD. i. 3.

On DE describe the equilateral triangle DFE. i. 1.
Join CF.
Then shall the straight line CF be at right angles to AB.

l^roof. For in the triangles DCF, ECF,

DC is equal to EC, Constr.

and CF is common to both ;
and the third side DF is equal to the third side
EF: Def. 19.

Therefore the angle DCF is equal to the angle ECF: i. 8.
But when a straight line, standing on another straight
line, makes the adjacent angles equal to one another, each
of these angles is called a right an^le; Def. 7.

therefore each of the angles DCF, ECF is a right angle.
Therefore CF is at right angles to AB,
and has been drawn from a point C in it. q.e.p.

EXERCISE.
In the figure of the above proposition, shew that amj point in
FC, or FC produced, is equidistant from D and E.

J

26 euclid s elements.

Proposition 12. Problem.

To draw a straight line perpendicular to a given straight
line of unlimited length, from a given point ivithotU it.

C

Let AB be tlie given straight line, which may be pro-
duced in either direction, and let C be the given point with-
out it.

It is required to draw from the point C a straight line
perpendicular to AB.

Construction. On the side of AB remote from C take
any point D;
and from centre C, with radius CD, describe the circle FDG,
meeting AB at F and G. Post. 3.

Bisect FG at H ; I. 10.

and join CH.
Then shall the straight line CH be perpendicular to AB. â–
Join CF and CG.
l^roof. Then in the triangles FHC, GHC,

( FH is equal to GH, Constr.

â– p I and HC is common to both;

[and the third side CF is equal to the third side

I CGj being radii of the circle FDG ; Def 11.

therefore the angle CHF is equal to the angle CHG; I. 8.

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

Therefore CH is a perpendicular drawn to the given
straight line AB from the given point C without it. q. e.f.

Note. The given straight line AB must be of unlimited length,
that is, it must be capable of production to an indefinite length in
either direction, to ensure its being intersected in two points by the
circle FDG.

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