Euclid.

# Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Font size and DCE the given rectilineal angle.
It is required, at the given point A in the given straight line AB,
make an angle that shall be equal to the given rectilineal angle DCJ

In CD, CE, take any points Z), E, and join DE',
on AB, make the triangle AFG, the sides of which shall be equal
to the three straight lines CE, EE, EC, so that AF be equal to
CE, AG to CE, and FG to EE. {i. 22.)

Then the angle FA G shall be equal to the angle E CE.

Because FA, AG are equal to EC, CE, each to each,

and the base FG is equal to the base EE ;

therefore the angle FAG is equal to the angle ECJEJ. (l. 8.)

Wherefore, at the given point A in the given straight line AB, the

angle FAG is made equal to the given rectilineal angle DCE. Q.E.F.

PROPOSITION XXIV. THEOREM.

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

Tuet ABC, EEFhe two triangles, which have the two sides AB,
A C, equal to the two EE, EF, each to each, namely, AB equal to
EE, and A Cto EF; but the angle 5^ C greater than the angle EEF.
Then the base BC shall be greater than the base EF.

BOOK I. PROP. XXIV, XXV. 23

IHr Of the two sides DE, DF, let DJS be not greater than JDF,

at the point 2), in the line DE, and on the same side of it as I)F,

make the angle FDG equal to the angle JBAC; (l. 23.)

make DG equal to DForAC, (i. 3.) and join EG, GF.

Then, because DE is equal to AB, and EG to AC,

the two sides EE, EG are equal to the two AE, AC, each to each,

and the angle EEG is equal to the angle BAC;

therefore the base EG is equal to the base EC. (l. 4.)

And because Z>^ is equal to EFm the triangle EFG,

therefore the angle EFG is equal to the angle EGF; (i. 5.)

but the angle EGF is greater than the angle EGF; (ax. 9.)

therefore the angle EFG is also greater than the angle EGF;

much more therefore is the angle EFG greater than the angle EGF.

And because in the triangle EFG, the angle EFG is greater than

the angle EGF,

and that the greater angle is subtended by the greater side ; (l. 19.)

therefore the side EG is greater than the side EF;

but EG was proved equal to EC;

therefore i?C is greater than EF.

"Wherefore, if two triangles, &c. Q.e.d.

PROPOSITION XXV. THEOREM.

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

Let ABC, EEFhe two triangles which have the two sides AB, AC,
equal to the two sides EE, EF, each to each, namely, AB equal to
EE, and ^Cto EF; but the base EC greater than the base EF.
Then the angle BA C shall be greater than the angle EEF.

For, if the angle BA Che not greater than the angle EEF,

it must either be equal to it, or less than it.

If the angle BA C were equal to the angle EEF,

then the base ^C would be equal to the ha.&e EF-, (l. 4.)

but it is not equal, (hyp.)

therefore the angle BAC is not equal to the angle EEF.

Again, if the angle BAC were less than the angle EEF,

then the base BC would be less than the base EF; (i. 24.)

but it is not less, (hyp.)

therefore the angle BA C is not less than the angle EEF;

mdithas been shewn, that the angle BA (7 is not equal to the angle EEF .

therefore the angle BA C is greater than the angle EEF.

Wherefore, if two ti'iangles, &c. Q. E. D.

24: Euclid's elements.

PROPOSITION XXYI. THEOREM.

If two triangles have two angles of the one equal to two angles of the
other, each to each, and owe side equal to one side, viz. either the sides adja-
cent to the equal angles in each, or the sides opposite to them ; then shall the
other sides be equal, each to each^ and also the third angle of the one equal
to the third angle of the other.

Let ABC, DUFhe two triangles which have the angles ABC,
BCA, equal to the angles DBF, EFB, each to each, namely, ABC
to BEF^ and BCA to EFD-, also one side equal to one side.

First, let those sides be equal which are adjacent to the angles that
are equal in the two triangles, namely, BC to EF.

Then the other sides shall be equal, each to each, namely, AB to
DE, and ^ C to DF, and the third angle BA C to the third angle EDF.

For, if ^^ be not equal to DE,
one of them must be greater than the other.

If possible, let AB be greater than DE,

make BG equal to ED, (l. 3.) and join GC.

Then in the two triangles GBC, DEF,

because GB is equal to DE, and ^Cto EF, (hyp.)

the two sides GB, BC&re equal to the two DE, EF, each to each ;

and the angle GBC is equal to the angle DEF;

therefore the base 6^Cis equal to the base DF, (i. 4.)

and the triangle GB C to the triangle DEF,

and the other angles to the other angles, each to each, to which

the equal sides are opposite ;

therefore the angle GCB is equal to the angle DFE;

but the angle A CB is, by the hypothesis, equal to the angle DFE ;

wherefore also the angle GCB is equal to the angle ACB; (ax. 1.)

the less angle equal to the greater, which is impossible ',

therefore AB is not unequal to DEf

that is, AB is equal to DE.

Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.)

and the angle ABC is equal to the angle DEF; (hyp.)

therefore the base ^ C is equal to the base DF, (i. 4.)

and the third angle BA C to the third angle EDF.

Secondly, let the sides which are opposite to one of the equal angles

in each triangle be equal to one another, namely, AB equal to DE.

Then in this case likewise the other sides shall be equal, AC to DF,
and ^ C to EF, and also the third angle BA C to the third angle EDF,

BOOK. I. PROP. XXVI, XXVII. 25

A D

n c
For if J5C be not equal to EF,
one of them must be greater than the other.

If possible, let ^Cbe greater than EF',
make jB7/ equal to EF, (l. 3.) and join AH.

Then in the two triangles ABH, DEF,

because AB is equal to DE, and BH to EF,

and the angle ABH to the angle JDEF; (hyp.)

therefore the base AH is equal to the base DF, (l. 4.)

and the triangle ABH to the triangle DEF,

and the other angles to the other angles, each to each, to which the

equal sides are opposite ;

therefore the angle BHA is equal to the angle EFD \

but the angle EFD is equal to the angle BCA ; (hyp.)

therefore the angle BHA is equal to the angle BCA, (ax. 1.)

that is, the exterior angle BHA of the triangle AHC, is

equal to its interior and opposite angle BCA ;

which is impossible ; (l. 16.)

wherefore BCh not unequal to EF^

that is, BCis equal to EF.

Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.)

and the included angle ABCis equal to the includedangle DEF; (hyp.)

therefore the base ^ C is equal to the base DF, (l. 4.)

and the third angle BA C to the third angle EDF.

Wherefore, if two triangles, &c. Q. E. d.

PROPOSITION XXVII. THEOREM.
If a straight line falling on two other straight lines, make the alternate
angles equal to each other ; these tioo straight lines shall be parallel.

Let the straight line EF, which falls upon the two straight lines
AB, CD, make the alternate angles AEF^ EFD, equal to one another.
Then AB shall be parallel to CD.

I

For, if ^2? be not parallel to CD,
then AB and CD being produced will meet, either towards A and C,
or towards B and D.
Tuet AB, CD be produced and meet, if possible, towards B and D,
in the point G, ^

then GEF is a triangle.

c

26 EUCL1D*S ELEMENTS.

And because a side GE of the triangle GEF is produced to A,
therefore its exterior angle AEF is greater than the interior and
opposite angle EFG ; (i. 16.)

but the angle AFFis equal to the angle FFG ; (hyp.)
therefore the angle AFF is greater than, and. equal to, the angle
FFG ; which is impossible.
Therefore AB, CD being produced, do not meet towards J5, D.
In like manner, it may be demonstrated, that they do not meet
when produced towards A, C,

But those straight lines in the same plane, which meet neither way,
though produced ever so far, are parallel to one another ; (def. 35.)
therefore AB h parallel to CD.
Wherefore, if a straight line, &c. q.e.d.

i

PROPOSITION XXVIII. THEOREM.

If a straight line falling upon two other straight lines ^ make the exterior
angle equal to the interior and opposite upon the same side of the line ; or
make the interior angles upon the same side together equal to two right
angles ; the two straight lines shall be parallel to one another.

Let the straight line FF, which falls upon the two straight lines
AB, CD, make the exterior angle FGB equal to the interior and
opposite angle GHD, upon the same side of the line FF-, or make
the two interior angles BGH, GHD on the same side together
equal to two right angles.

Then AB shall be parallel to CD.

B
g"
C -V D

Because the angle EGB is equal to the angle GHD, (hyp.)
and the angle EGB is equal to the angle A GH, (i. 15.)
therefore the angle A GH is equal to the angle GHD; (ax. 1.)
and they are alternate angles,

therefore AB is parallel to CD. (l. 27.)
Again, because the angles BGH, GHD are together equal to t\
right angles, (hyp.)

and that the angles AGH, BGH ÂŁire also together equal to twl

right angles; (l. 13.)
therefore the angles AGH, BGH are equal to the angles BGj
GHD', (ax. 1.)

take away from these equals, the common angle BGH;
therefore the remaining angle A GH is equal to the remaining angl
GHD; (ax. 3.)

and they are alternate angles ;
therefore AB is parallel to CD. (l. 27.)
"Wherefore, if a straight line, &c. Q. E. D.

BOOK I. PROP. XXIX.

PROPOSITION XXIX. THEOREM.

Jf a straight line fall upon two parallel straight lines, it makes the alter-
nate angles equal to one another ; and the exterior angle equal to the interior
and opposite upon the satne side ; and likewise the two interior angles upo?i
the same side together equal to two right angles.

Let the straight line ^jPfall upon the parallel straight lines AB, CD.
Then the alternate angles A Gil, GIID shall be equal to one another ;

the exterior angle IJGB shall be equal to the interior and opposite
angle GUI) upon the same side of the line UF;
and the two interior angles BGH, GIID upon the same side of JSF

shall be together equal to two right angles.

B

C 77 V D

First. For, if the angle A GH be not equal to the alternate angle

I GIID, one of them must be greater than the other ;

if possible, let A Gil be greater than GHD,

then because the angle A GH is greater than the angle GHD,

add to each of these unequals the angle BGH;

therefore the angles AGH, BGH are greater than the angles BGH,

GIID', (ax. 4.)
but the angles AGH, BGH are equal to two right angles ; (l. 13.)
therefore the angles BGH, GHD are less than two right angles ;
but those straight lines, which with another straight line falling upon
them, make the two interior angles on the same side less than two
right angles, will meet together if continually produced; (ax. 12.)
therefore the straight lines AB, CD, if produced far enough, will
meet towards B, D ;
but they never meet, since they are parallel by the hypothesis ;

therefore the angle A GH is not unequal to the angle GHD,

that is, the angle AGH is equal to the alternate angle GHD.

Secondly, because the angle ^6^^ is equal to the angle JEGB, (i. 15.)

and the angle A GH is equal to the angle GHD,

therefore the exterior angle JEGB is equal to the interior and opposite

angle GIID, on the same side of the line.

Thirdly. Because the angle DGB is equal to the angle GHD,

add to each of them the angle BGH;

therefore the angles FGB, BGH axe equal to the angles BGH, GHD ;

(ax. 2.)

but EGB, BGH axe equal to two right angles ; (l. 13.)
therefore also the two interior angles BGH, GIID on the same side
of the line are equal to two right angles, (ax. 1.)
Wherefore, if a straight line, &c. q.e.d.

c2

^8 Euclid's elements.

PROPOSITION XXX. THEOREM.

Straight lines which are parallel to the same straight line are parallel to
each other.

Let the straight lines AB, CD, be each of them parallel to EF.
Then shall AB be also parallel to CD.

A ^B

E -/- F

C /^ D

Let the straight line GHK cut AB, EF, CD.
Then because GHK cuts the parallel straight lines AB, EF, in
a, II:

therefore the angle A GH is equal to the alternate angle GHF. (l. 29.)
Again, because GHK cuts the parallel straight lines EF, CD, in
H,K',

therefore the exterior angle GHF is equal to the interior angle HKD ;

and it was shewn that the angle AGIIh equal to the angle GHF;

therefore the angle AGHis equal to the angle GKD;

and these are alternate angles ;

therefore ^i? is parallel to CD, (l. 27.)

Wherefore, straight lines which are parallel, &c. q.e.d.

PROPOSITION XXXI. PROBLEM.

To draw a straight line through a given point parallel to a giveti straight
line.

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

E A F

In the line J? C take any point D, and join AD;
at the point A in the straight line AD,
make the angle DAE equal to the angle ADC, (i. 23.) on the oppo

and produce the straight line EA to F.

Then EF shall be parallel to BC.

Because the straight line AD meets the two straight lines EF, BC

therefore ^i^is parallel to BC (l. 27.)

Wherefore, through the given point A, has been drawn a straigh

line ÂŁ^jP parallel to the given straight line BC. Q.E.F.

BOOK 1. PROP. XXXll. 29

PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the
two interior and opposite angles ; and the three interior angles of every
triangle are together equal to two right angles.

Let ^i?(7be a triangle, and let one of its sides ^ C be produced to Z).

Then the exterior angle A CD shall be equal to the two interior
and opposite angles CAB, ABC:

and the three interior angles ABC, BCA, CAB shall be equal to
two right angles.

A E

Through the point C draw CE parallel to the side BA. (l. 31.)
Then because CE is parallel to BA, and A C meets them,
therefore the angle A CE is equal to the alternate angle BA C (l. 29.)
Again, because CE is parallel to AB, and BD falls upon them,
therefore the exterior angle ECD is equal to the interior and op-
posite angle ABC; (l. 29.)
but the angle ACE was shewn to be equal to the angle BAC;
therefore the whole exterior angle A CE is equal to the two interior
and opposite angles CAB, ABC: (ax. 2.)
Again, because the angle A CE is equal to the two angles ABC, BA C,
to each of these equals add the angle ACB,
therefore the angles A CE and A CB are equal to the three angles

ABC, BAC, and A CB. (ax. 2.)
but the angles ACE, ACB are equal to two right angles, (l. 13.)
therefore also the angles ABC, BAC, ACB are equal to two right

angles, (ax. 1.)
Wherefore, if a side of any triangle be produced, &c. Q. E D.
Cor. 1. All the interior angles of any rectilineal figure together
with four right angles, are equal to twice as many right angles as the
figure has sides.

D

For any rectilineal figure ABCEE can be divided into as many
triangles as the figure has sides, by drawing straight lines from a point
F within the figure to each of its angles.

â€˘ Then, because the three interior angles of a triangle are equal to

two right angles, and there are as many triangles as the figure has sides,

therefore all the angles of these triangles are equal to twice as many

right angles as the figure has sides ;
but the same angles of these triangles are equal to the interior angles
of the figure together with the angles at the point JP:

30 Euclid's elements.

and the angles at the point F, which is the common vertex of all

the triangles, are equal to four right angles, (l. 15. Cor. 2.)
therefore the same angles of these triangles are equal to the angles

of the figure together with four right angles ;
but it has been proved that the angles of the triangles are equal to

twice as many right angles as the figure has sides ;
therefore all the angles of the figure together with four right angles,

are equal to twice as many right angles as the figure has sides.
Cor. 2. All the exterior angles of any rectilineal figure, made by
producing the sides successively in the same direction, are together
equal to four right angles.

Since every interior angle ABCvf\i\i its adjacent exterior angle
ABD, is equal to two right angles, (l. 13.)

therefore all the interior angles, together with all the exterior angles,
are equal to twice as many right angles as the figure has sides ;

but it has been proved by the foregoing corollary, that all the in-
terior angles together with four right angles are equal to twice as many
right angles as the figure has sides ;

therefore all the interior angles together with all the exterior angles,
are equal to all the interior angles and four right angles, (ax. 1.)
take from these equals all the interior angles,

therefore all the exterior angles of the figure are equal to four right
angles, (ax. 3.)

PROPOSITION XXXIII. THEOREM.

The straight lines ichich join the extremities of two eqxial and parallel
straight lines totcards the same parts, are also themselves equal and parallel.

Let AB, CD be equal and parallel' straight lines,

and joined towards the same parts by the straight lines A C, BD.

Then A C, BD shall be equal and parallel.

Join BC.

Then because ^.5 is parallel to CD, and ^C meets them,

therefore the angle ABCis equal to the alternate angle BCD ; (l. 29.)

and because ^5 is equal to CD, and ^C common to the two triangles

ABQDCB; the two sides AB, BC, are equal to the two DC, CB, each

to each, and the angle ABC^vas proved to be equal to the angle jB CD:

therefore the base A C is equal to the base BD, (i. 4.)

and the triangle ABC to the triangle BCD,

t

r

^^nd the oth(

BOOK I. PROP. XXXIV. 31

md the other angles to the other angles^ each to each, to which the
equal sides are opposite ;
therefore the angle A CB is equal to the angle CJBD.
And because the straight line ^C meets the two straight lines AC,
BD, and makes the alternate angles A CB, CBD equal to one another ;
therefore ACk parallel to BJD ; (i. 27.)
and A C was shewn to be equal to BJD.
Therefore, straight lines which, &c. Q.E.D.

PROPOSITION XXXIV. THEOREM.

The opposite sides and angles of a parallelogram are equal to one another,
and the diameter bisects it, that is, divides it into two equal parts.

Let A CDB be a parallelogram, of which ^C is a diameter.
Then the opposite sides and angles of the figure shall be equal to
one another ; and the diameter J5C shall bisect it.

/

y

Because AB is parallel to CD, and ^C meets them,
therefore the angle ^-SC is equal to the alternate angle BCD. (l. 29.)

And because ^ C is parallel to BD, and i?C meets them,
therefore the angle ACB is equal to the alternate angle CBD. (l. 29.)

Hence in the two triangles ABC, CBD,
because the two angles ABC, BCA in the one, are equal to the two

angles BCD, CBD in the other, each to each;
and one side BC, which is adjacent to their equal angles, common to

the two triangles ;
therefore their other sides are equal, each to each, and the third angle
of the one to the third angle of the other, (l. 26.)
namely, the side AB to the side CD, and AC to BD, and the angle
B AC to the angle BDC.

And because the angle ABC h equal to the angle BCD,
and the angle CBD to the angle A CB,
therefore the whole angle ABD is equal to the whole angle A CD ;
(ax. 2.)

and the angle J5^ Chas been shewn to be equal to BDC;
therefore the opposite sides and angles of a parallelogram are equal to
one another.
Also the diameter JBC bisects it.

For since AB is equal to CD, and ^(7 common, the two sides AB,

BC, are equal to the two DC, CB, each to each,

and the angle ^.SC has been proved to be equal to the angle BCD ;

therefore the triangle ABC i^ equal to the triangle BCD : (l. 4.) and

the diameter ^C divides the parallelogram A CDB into two equal parts.

Q.E.D.

S2 Euclid's elements.

PROPOSITION XXXV. THEOREM.

Parallelograms upon the same base^ and between the same parallels, are
equal to one another.

Let the parallelograms ^5 CD, ^^ OF be upon the same base J5C;
and between the same parallels AF, BC.

Then the parallelogram ^i? CD shall be equal to the parallelogram
JESCF.

A E D F

If the sides AD, DJ'of the parallelograms A BCD, DBCF, opposite
to the base BC, be terminated in the same point D ;
then it is plain that each of the parallelograms is double of the triangle

BDC', (I. 34.)
and therefore the parallelogram A BCD is equal to the parallelogram
DBCF. (ax. 6.)

But if the sides AD, JEF, opposite to the base BC, be not termi-
nated in the same point ;

Then, because A BCD is a parallelogram,

therefore AD is equal to BC; (l. 34.)
and for a similar reason, FF is equal to BC;
wherefore ^D is equal to FF; (ax. 1.)
and DF is common ;
therefore the whole, or the remainder AF, is equal to the whole, or
the remainder DF ; (ax. 2 or 3.)

and AB is equal to DC; (l. 24.)
hence in the triangles FAB, FDC,
because FD is equal to FA, and DC to AB,
and the exterior angle FDC is equal to the interior and opposite angle
FAB ; (I. 29.)

therefore the base FC is equal to the base FB, (l. 4.)
and the triangle FDCis equal to the triangle FAB.
From the trapezium ABCF take the triangle FDC,
and from the same trapezium take the triangle FAB,
and the remainders are equal, (ax. 3.)
therefore the parallelogram^^ CD is equal to the parallelogram^^? Ci^.
Therefore, parallelograms upon the same, &c. Q. E. D.

PROPOSITION XXXVI. THEOREM.

Parallelograms upon equal bases and between the same parallels, are
equal to one another.

Let ABCD, FFGJThe parallelograms upon equal bases BC, FG,
and between the same parallels AH, BG.

Then the parallelogram AB CD shall be equal to the parallelogram
FFGÂŁ[.

BOOK I. PROP. XXXVI, XXXVII. 33

A D E

Join BE, CH.
Then because ^Cis equal to FG, (hyp.) and FG to EH, (l. 34.)
therefore JBCis equal to EIT; (ax, 1.)
and these lines are parallels, and joined towards the same parts by the
straight lines EE, CH-,
but straight lines which join the extremities of equal and parallel
straight lines towards the same parts, are themselves equal and parallel ;
(I. 33.)

therefore BE, CH are both equal and parallel ;
wherefore EBCHis a parallelogram, (def. A.)
And because the parallelograms ABCD, EBCH, are upon the

same base BC, and between the same parallels BC, AH',
therefore the parallelogram ABCD is equal to the parallelogram
EBCH. (1.35.)

For the same reason, the parallelogram EFGH is equal to the
parallelogram EBCH;

therefore the parallelogram ABCD is equal to the parallelogram
EFGH (ax. 1.)

Therefore, parallelograms upon equal, &c. q.e.d.

PHOPOSITION XXXVII. THEOREM.

Triangles upon the same base a7id between the same parallels, are equal to
one another.

Let the triangles ABC, DBChe upon the same base BCf
and between the same parallels AD, BC.
Then the triangle ABC shall be equal to the triangle DBC.
E A D F

Produce AD both ways to the points E, F;

through B draw BE parallel to CA, (l. 31.)

and through (7 draw CF parallel to BD.

Then each of the figures EBCA, DBCF is a parallelogram ;

md EBCA is equal to DBCF, (i. 35.) because they are upon the

same base BC, and between the same parallels BC, EF.

And because the diameter AB bisects the parallelogram EBCA,

therefore the triangle ABC is half of the parallelogram EBCA ; (i. 34.)

also because the diameter i)C bisects the parallelogram DBCF,

therefore the triangle DBC is half of the parallelogram DBCF,

but the halves of equal things are equal ; (ax. 7.)

therefore the triangle ABC is equal to the triangle DBC.

Wherefore, triangles, &c. Q.E.D.

C5

PROPOSITION XXXVIII. THEOREM.

Triangles upon equal bases and between the same parallels, are equal
to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and
between the same parallels BF, AD.

llien the triangle AB C shall be equal to the triangle DEF.
G A D n

Produce AD both ways to the points G, IT;

through B draw BG parallel to CA, (i. 31.)

and through F draw FH parallel to ED.