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

. (page 2 of 38)
Font size if the base BC diO not coincide with the base EF, the two straight lines
i?Cand JS'i^ would enclose a space, which is impossible, (ax. 10.)
Therefore the base BC does coincide with EF, and is equal to it;
and the whole triangle ABC coincides with the whole triangle
DEF, and is equal to it ;

also the remaining angles of one triangle coincide with the remain-
ing angles of the other, and are equal to them,

viz. the angle ABC to the angle DEF,
and the angle A CB to DFE.
Therefore, if two triangles have two sides of the one equal to two
sides, &c. "Which was to be demonstrated.

PROPOSITION V. THEOREM.

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

Let ABCheoii isosceles triangle of which the side AB is equal to A C,

and let the equal sides AB, AChe produced to D and E.

Then the angle ^J5C shall be equal to the angle ACB,

and the angle DBC to the angle ECB.

In JBD take any point F;

from AE the greater, cut oQ AG equal to AF the less, (l. 3.)

and join EC, GB.

Because AF\^ equal to AG, (constr.) and AB to AC-, (hyp.)

the two sides FA, ^Care equal to the two GA, AB, each to each;

and they contain the angle FA G common to the two triangles

AFCAGB; /

b5

10 Euclid's elements.

therefore the base FC is equal to the base GB, (l. 4.)

and the triangle AFC is equal to the triangle A GB,

also the remaining angles of the one are equal to the remaining angles

of the other, each to each, to which ;the equal sides are opposite ;

viz. the angle A CF to the angle ABG,

and the angle AFC to the angle A GB.

And because the whole AFis equal to the whole AG,

of which the parts AB, AC, are equal;

therefore the remainder BF is equal to the remainder CG ; (ax. 3.)

and FC has been proved to be equal to GB ;

hence, because the two sides BF, FC are equal to the two CG, GB,

each to each ;
and the angle BFChas been proved to be equal to the angle CGB,
also the base BCh common to the two triangles BFC, CGB;
wherefore these triangles are equal, (l. 4.)
and their remaining angles, each to each, to which the equal sides
are opposite ;

therefore the angle FBCh equal to the angle GCB,
and the angle BCF to the angle CBG.
And, since it has been demonstrated,

that the whole angle ABG is equal to the whole ACF,

the parts of which, the angles CBG, BCF are also equal;

therefore the remaining angle ABCk equal to the remaining angle A CB,

which are the angles at the base of the triangle ABC;

and it has also been proved,

that the angle FBCk equal to the angle GCB,

which are the angles u^on the other side of the base.

Therefore the angles at the base, &c. q.e.d.

Cor. Hence an equilateral triangle is also equiangular.

PROPOSITION VI. THEOREM.

Jf txco angles of a triangle be equal to each other ; the sides also which
subtend, or are opposite to, the equal angles, shall be equal to one another.

Let ABChe a triangle having the angle ABCequol to the angle A CB.
Then the side AB shall be equal to the side A C.

BOOK I. PROP. VI, VII. 11

For, if AJB be not equal to A C,

one of them is greater than the other.

If possible, let AB be greater than AC;

and from BA cut off ^i) equal to Cui the less, (l. 3.) and join DC.

Then, in the triangles DBC, ABC,

because BB is equal to A C, and ^C is common to both triangles,

the two sides JOB, BCare equal to the two sides A C, CB, each to each ;

and the angle BBCis equal to the angle A CB ; (hyp.)

therefore the base JDCk equal to the base AB, (l. 4.)

and the triangle DBCis equal to the triangle ABC,

the less equal to the greater, which is absurd, (ax. 9.)

Therefore AB is not unequal to A C, that is, AB is equal to AC.

Wherefore, if two angles, &c. Q.E.D.

Cor. Hence an equiangular triangle is also equilateral.

PROPOSITION VII. THEOREM.

Upon the same base, and on the same side of it, there cannot he two
triangles that have their sides which are terminated in one extremity of the
base, equal to one another, and likewise those which are terminated in the
other extremity.

If it be possible, on the same base AB, and upon the same side of
it, let there be two triangles ACB, ADB, which have their sides CA,
DA, terminated in the extremity A of the base, equal to one another,
and likewise their sides CB, DB, that are terminated in B.

C D

Join CD.

First. When the vertex of each of the triangles is without the
other triangle.

Because ACis equal to AD in the triangle A CD,
therefore the angle ADC is equal to the angle A CD; (l. 5.)
but the angle A CD is greater than the angle BCD ; (ax. 9.)

therefore also the angle ADC is greater than BCD;

much more therefore is the angle jl5^ (7 greater than BCD.

Again, because the side j^Cis equal to BD in the triangle BCD, (hyp.)

therefore the angle BDCis equal to the angle BCD ; (i. 5.)

but the angle BDCwecs proved greater than the angle BCD,

hence the angle BDCis both equal to, and greater than the angle J? CD;

which is impossible.
Secondlv- Let the vertex D of the triangle ADB fall within the
triangle ..4 OB.

12

Produce AC to H, and AD to F, and join CD.
Then because ^ C is equal to AD in the triangle A CD,
therefore the angles JECD, FDC u^on the other side of the base CD,
are equal to one another ; (i. 5.)

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

therefore also the angle FDCh greater than the angle BCD ;

much more then is the angle J5Z) 5 greater than the angle BCD.

Again, because ^Cis equal to BD in the triangle BCD-,

therefore the angle BDCh equal to the angle BCD, (i. 5.)

but the angle ^BDChas been proved greater than BCD,

wherefore the angle BDCh both equal to, and greater than the

angle BCD ; which is impossible.

Thirdly. The case in which the vertex of one triangle is upon a
side of the other, needs no demonstration.
Therefore, upon the same base and on the same side of it, &c. q.e.d.

PROPOSITION VIII. THEOREM.

Tf two triangles have two sides of the one equal to two sides of the other,
each to each, a7id have liketoise their bases equal; the angle which is con-
tained by the tico sides of the one shall be equal to the angle contai^ied by the
txco sides equal to them, of the other.

Let ABC, DFFhe two triangles, having the two sides AB, AC,
equal to the two sides DF, DF, each to each, viz. AB to DF, and
^ C to DF, and also the base J5C equal to the base FF.

A D G

Then the angle BA C shall be equal to the angle EDF.
For, if the triangle ABC he applied to DFF,
so that the point B be on E, and the straight line BC on FF;
then because BC is equal to FF, (hyp.)
therefore the point C shall coincide with the point F;
wherefore ^C coinciding with FF,
BA and A C shall coincide with FD, DF;
for, if the base jBC coincide with the base FF, but the sides BA, A C,
do not coincide with the sides FD, DF, but have a different situation
as FG, GF:

then, upon the same base, and upon the same side of it, there can

be two triangles which have their sides wliich are terminated in one

extremity of the base, equal to one another, and likewise those sides

which are terminated in the other extremity; but this is impossible. (l. 7.)

Therefore, if the base BC coincide with the base FF,

the sides BA, ^C cannot but coincide with the sides FD, DF;

wherefore likewise the angle BA C coincides with the angle FDF, and

is equal to it. (ax. 8.)

Therefore if two triangles have two sides, &c. q.e.d.

BOOK I. PROP. IX, X. 13

I PROPOSITION IX. PROBLEM.

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

Let BA C be the given rectilineal angle.
It is required to bisect it.

In AB take any point D ;

from A C cut off AE equal to AD, (i. 3.) and join BE ;

on the side of DE remote from A,

describe the equilateral triannjle DEF(l. 1), and join AF,

Then the straight line ^-F shall bisect the angle BAC,

Because AD is equal to AE, (constr.)

and AF\^ common to the two triangles DAF, EAF;

the two sides DA, AF, are equal to the two sides EA, AF, each to each;

and the base Di^is equal to the base EF: (constr.)

therefore the angle DAF is equal to the angle EAF. (I. 8.)

Wherefore the angle BA C is bisected by the straight line AF. Q.e.f.

PROPOSITION X. PROBLEM.

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

Let ABhe the given straight line.
It is required to divide AB into two equal parts.
Upon AB describe the equilateral triangle ABC; (I. 1.)

and bisect the angle A CB by the straight line CD meeting AB in the
point Z>. (1.9.)

Then AB shall be cut into two equal parts in the point D,

Because ^Cis equal to CB, (constr.)

and CD is common to the two triangles A CD, BCD ;

the two sides AC, CD are equal to the two BC, CD, each to each;

and the angle A CD is equal to BCD; (constr.)

therefore the base AD is equal to the base BD. (I. 4.)

Wherefore the straight line AB is divided into two equal parts in the

point D. Q.E.F.

L

14 Euclid's elements.

PROPOSITION XI. PROBLEM.

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

Let AJB be the given straight line, and C a given point in it.
It is required to draw a straight line from the point C at right
angles to AJ3.

F

In retake any point D, and make CJE equal to CD ; (l. 3.)
upon DU describe the equilateral triangle DUF (i. 1), and join CF.
Then CF drawn from the point C, shall be at right angles to AJB.
Because DCis equal to FC, and FCis common to the two triangles
DCF, FCF;
the two sides DC, CF are equal to the two sides FC, CF, each to each ;
and the base .DFis equal to the base FF; (constr.)
therefore the angle FCF is equal to the anglfe FCF: (i. 8.)
and these two angles are adjacent angles.
But when the two adjacent angles which one straight line makes
with another straight line, are equal to one another, each of them is
called a right angle : (def. 10.)

therefore each of the angles FCF, FCF is a right angle.
"V^Tierefore from the given point C, in the given straight line AB,
FC has been drawn at right angles to AB. Q-E.F.

Cor. By help of this problem, it may be demonstrated that two
straight lines cannot have a common segment.

If it be possible, let the segment ABhe common to the two straight
lines ABC, ABB.

ABC

From the point B, draw BF at right angles to AB; (T. 11.)

then because ABC is a straight line,

therefore the angle ABF is equal to the angle FBC, (def. 10.)

Similarly, because ABF is a straight line,

therefore the angle ABF is equal to the angle FBD;

but the angle ABF is equal to the angle FBC,

wherefore the angle FBD is equal to the angle FBC, (ax. 1.)

the less equal to the greater angle, which is impossible.
Therefore two straight lines cannot have a common segment.

PROPOSITION XII. PROBLEM.

To draw a straight line perpetidicular to a given straight litie of un-
limited length, from a given pohit without it.

BOOK I. PROP. XII, XIII. 15

Let AB he the given straight line, which may be produced any
length both ways, and let C be a point without it.

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

Upon the other side oi AB take any point D,
and from the center C, at the distance CD, describe the circle EGF
meeting AB, produced if necessary, in i^and G: (post. 3.)
bisect FG in II(i. 10.), and join CH.
Then the straight line CH drawn from the given point C, shall be
perpendicular to the given straight line AB.
Join FC, and CG.
Because FHis equal to JIG, (constr.)
and HCis common to the triangles FHC, GUC',
the two sides FH, HC, are equal to the two GH, JSC, each to each ;
and the base Ci^is equal to the base CG', (def. 15.)
therefore the angle FHC is equal to the angle GHC; (l. 8.)
But when a straight line standing on another straight line, makes
the adjacent angles equal to one another, each of them is a right angle,
and the straight line which stands upon the other is called a perpen-
dicular to it. (def. 10.)

Therefore from the given point C, a perpendicular CH has been
wn to the given straight line AB. q.e.f.

w

PROPOSITION XIII. THEOREM.

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

Let the straight line AB make with CD, upon one side of it, the
angles CBA, ABD.

Then these shall be either two right angles,
or, shall be together, equal to two right angles.

E
A A

For if the angle CBA be equal to the angle ABD,

each of them is a right angle, (def. 10.)

But if the angle CBA be not equal to the angle ABD,

from the point B draw BE at right angles to CD. (I. 11.)

Then the angles CBE, EBD are two right angles, (def. 10.)

16

And because the angle CBE is equal to the angles CBA, ABE,

add the angle EBD to each of these equals ;

therefore the angles CBE, EBD are equal to the three angles CBa,

ABE, EBD. (ax. 2.)
Again, because the angle DBA is equal to the two angles DBE, EBA,

add to each of these equals the angle ABC;
therefore the angles DBA, ABC ave equal to the three angles DBE,
EBA, ABC.
But the angles CBE, EBD have been proved equal to the same
three angles ;

and things which are equal to the same thing are equal to one another ;
therefore the angles CBE, EBD are equal to the angles DBA, ABC-,

but the angles CBE, EBD are two right angles ;
therefore the angles DBA, ABC sue together equal to two right angles,
(ax. 1.)

Wherefore, when a straight line, &c. Q.E.D.

PROPOSITION XIV. THEOREM.

If at a point in a straight line, two other straight lines, upon the opposite
sides of it, make the adjacent angles together equal to two right angles ; t/ien
these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines
BC, BD upon the opposite sides of ^5, make the adjacent angles
ABC, ABD together equal to two right angles.

Then BD shall be in the same straight line with BC.

For, if BD be not in the same straight line with BC,
if possible, let BE be in the same straight line with it.
Then because AB meets the straight line CBE-,
therefore the adjacent angles CBA, ABE are equal to two right angles ;

but the angles CBA, ABD are equal to two right angles ; (hyp.)
therefore the angles CBA, ABE are equal to the angles CBA, ABD :
(ax. 1.)
take away from these equals the common angle CBA,
therefore the remaining angle ABE is equal to the remaining angle
ABD ; (ax. 3.)
the less angle equal to the greater, which is impossible :
therefore BE is not in the same straight line with BC.
And in the same manner it may be demonstrated, that no other
can be in the same straight line with it but BD, which therefore is in
the same straight line with B C.

Wherefore, if at a point, &c. Q. e. b.

BOOK I. PROP. XV, XVT. 17

PROPOSITION XV. THEOREM.

If two straight lines cut one another ^ the vertical^ or opposite anr/ka
shall be equal.

Let the two straight lines AJB, CD cut one another in the point JE.

Then the angle AJEC shall be equal to the angle DEB, and the
angle CEB to the angle AED,

I

Because the straight line AE makes with CD at the point E, the

these angles are together equal to two right angles. (l. 13.)
Again, because the straight line DB makes with AB dX the point E,
the adjacent angles AED, DEB ;

these angles also are equal to two right angles ;
but the angles CEA, AED have been shewn to be equal to two right

angles ;
wherefore the angles CEA, AED are equal to the angles AED, DEB ;

take away from each the common angle AED,
and the remaining angle CEA is equal to the remaining angle DEB,
(ax. 3.)

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

Therefore, if two straight lines cut one another, &c. Q. E. D.
Cor. 1. From this it is manifest, that, if two straight lines cut each
other, the angles which they make at the point where they cut, are
together equal to four right angles.

Cor. 2. And consequently that all the angles made by any num-
ber of lines meeting in one point, are together equal to four right
angles.

PROPOSITION XVI. THEOREM.

If one side of a triangle he produced, the exterior angle is greater than
either of the hiterior opposite angles.

Let ABC he a triangle, and let the side BChe produced to D.
Then the exterior angle A CD shall be greater than either of ths
Jnterior opposite angles CBA or BA C.

Bisect ^C in E, (l. 10.) and join BE;
produce BE to F, making ^i^ equal to BE, (l. 3.) and join FC.

18

Because AE is equal to EC, and JBJ3 to EF; (coiistr.)
the two sides AU, EB are equal to the two CE, EF, each to each, in
the triangles AJBE, CFE;

and the angle AEB is equal to the angle CEF,

because they are opposite vertical angles; (i. 15.)

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

and the triangle AEB to the triangle CEF,

and the remaining angles of one triangle to the remaining angles of

the other, each to each, to which the equal sides are opposite ;

wherefore the angle JBAE is equal to the angle ECF;
but the angle ECD or ACE is greater than the angle ECF;
therefore the angle ACE is greater than the angle BAE or BAC.
In the same manner, if the side BC he bisected, and A Che pro-
duced to G^; it may be demonstrated that the angle J3CG, that is, the
angle ACE, (i. 15.) is greater than the angle ABC.

Therefore, if one side of a triangle, &c. Q. E. D.

PROPOSITION XVII. THEOREM.

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

Let ABC he any triangle.
Then any two of its angles together shall be less than two right angles.

Produce any side BC to D.
Then because A CE is the exterior angle of the triangle ABC;
therefore the angle A CD is greater than the interior and opposite angle
ABC; (I. 16.)

to each of these unequals add the angle A CB ;
therefore the angles ^CZ), ACB are greater than the angles ABC,
ACB;

but the angles A CD, ACB are eqaal to two right angles ; (i. 13.)

therefore the angles ABC, A CB are less than two right angles.

In like manner it may be demonstrated,

that the angles BA C, A CB are less than two right angles,

as also the angles CAB, ABC.

Therefore any two angles of a triangle, &c. Q. E. D.

PROPOSITION XVIII. THEOREM.

The greater side of every triangle is opposite to the greater angle.

Let ABC he a triangle, of which the side ^C is greater than the
side AB-

BOOK I. PROP. XVIIl â€” XX. 19

Then the angle ABC shall be greater than the angle ACB.

Since the side ACi^ greater than the side AB, (hyp.)

make AD equal to AB, (l. 3.) and join BD.

Then, because AD is equal to AB, in the triangle ABD,

therefore the angle ABD is equal to the angle ADB, (l. 5.)

but because the side CD of the triangle BDC is produced to A,

therefore the exterior angle ADB is greater than the interior and

opposite angle DCB; (l. 16.)

but the angle ADB has been proved equal to the angle ABD,

therefore the angle ABD is greater than the angle DCB ;

v/herefore much more is the angle ABC greater than the angle ACB.

Therefore the greater side, &c. Q. E. D.

PROPOSITION XIX. THEOREM.

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

Let ABChe B, triangle of which the angle ABCis gi-eater than the
angle BCA.

Then the side .4 C shall be greater than the side AB.

For, if ^ C be not greater than AB,

A C must either be equal to, or less than AB ;

if AC were equal to AB,

then the angle ABC would be equal to the angle ACB; (l. 5.)

but it is not equal ; (hyp.)

therefore the side ^ Cis not equal to AB.

Again, if ^ C were less than AB,

then the angle ^j^C would be less than the angle ACB; (l. 18.)

but it is not less, (hyp.)

therefore the side AC is not less than AB ;

and A C has been shewn to be not equal to AB ;

therefore AC is greater than AB.

Wherefore the greater angle, &c. Q. E. D.

, PROPOSITION XX. THEOREM.

Any two sides of a triangle are together greater than the third side.

Let ABChe a triangle.

Then any two sides of it together shall be greater than the thii'd side,

viz. the sides BA, A C greater than the side BC;

20

AB, JBC greater than AC;
and ÂŁC, CA greater than AB

Produce the side BA to the point D,

make AD equal to AC, (i. 3.) and join DC.

Then because ^D is equal to A C, (constr.)

therefore the angle A CD is equal to the angle ADC; (l. 5.J

but the angle BCD is greater than the angle A CD ; (ax. 9.)

therefore also the angle BCD is greater than the angle ADC,

And because in the triangle DBC,

the angle BCD is greater than the angle BDC,

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

therefore the side DB is greater than the side BC;

but DB is equal to BA and A C,

therefore the sides BA and AC are greater than BC

In the same manner it may be demonstrated,

that the sides AB, BCare greater than CA j

also that BC, CA are greater than AB,

Therefore any two sides, &c. Q. E. D.

PROPOSITION XXI. THEOREM.

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

Let ABC he a triangle, and from the points B, C, the ends of the
side B C, let the two straight lines BD, CD be drawn to a point D
within the triangle.

Then BD and DC shall be less than BA and AC the other two

sides of the triangle,
but shall contain an angle ^i)C greater than the angle BAC.

Produce BD to meet the side AC in U.
Because two sides of a triangle are greater than the third side, (l. 20.)
therefore the two sides BA, AE of the triangle ABE are greater
than BE;

to each of these unequals add EC;
therefore the sides BA, AC are greater than BE, EC. (ax. 4.)
Again, because the two sides CE, ED of the triangle CED are
greater than DC; (i. 20.)

add DB to each of these unequals ;

BOOK I. PROP. XXI, XXII. 21

erefore the sides CE, EB are greater than CD, DB. (ax. 4.)
But it has been shewn that BA, A C are greater than BE, EC)

much more then are BA, A C greater than BD, DC.
Again, because the exterior angle of a triangle is greater than the
interior and opposite angle ; (l. 16.)

therefore the exterior angle ^DCof the triangle CDE is greater
than the interior and opposite angle CED ;

for the same reason, the exterior angle CED of the triangle A BE
is greater than the interior and opposite angle BA C;
and it has been demonstrated,

that the angle BDCis greater than the angle CEB ;

much more therefore is the angle BDC greater than the angle BA C.

Therefore, if from the ends of the side, &c. q.e.d.

PROPOSITION XXII. PROBLEM.

To make a triangle of which the sides shall be equal to three given
straight lines, but any two whatever of these must be greater than the third.

Let A, B, C be the three given straight lines,

of which any two whatever are greater than the third, (l. 20.)

namely, A and B greater than (7;

A and C greater than B ;

and B and C greater than A.

It is required to make a triangle of which the sides shall be equal

to ^, B, C, each to each.

K

Take a straight line DE terminated at the point D, but unlimited
towards E,

make Z)-F equal to A, EG equal to B, and G^JT equal to C; (l. 3.)
from the center F, at the distance ED, describe the circle DKL',
(post 3.)

from the center G, at the distance GK, describe the circle SLK',
from X where the circles cut each other, di-aw KF, KG to the points
F,G',

Then the triangle KEG shall have its sides equal to the three
straight lines A, B, C.

Because the point F is the center of the circle DKL^

, therefore ED is equal to EK; (def. 15.)

but ED is equal to the straight line A ;

therefore EK is equal to A.

Again, because G is the center of the circle HKL ;

therefore Gllis equal to GK, (def. 15.)

but 6^// is equal to C;

therefore also GK is equal to C; (ax. 1.)

and EG is equal to B ;

22 Euclid's elements.

therefore the three straight lines XF, FG, GK, are respective!}-

equal to the three, A, B, C:
and therefore the triangle KFG has its three sides KF, FG, GK,

equal to the three given straight lines A, B, C. Q.E.r.

PROPOSITION XXIII. PROBLEM.

At a given jjoiiit in a given straight li7ie, to make a rectilineal angle
equal to a given rectilineal angle.

Let ^-S be the given straight line, and A the given point in it,