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 therefore, because the angle DAB is equal to the angle ABC,
(I. def. 30.)
and that the angle HAB is the half of DAB, and DBA the half of ^^ C;
therefore the angle DAB is equal to the angle DBA ; (ax. 7.)

wherefore the side DA is equal to the side DB : (l. 6.)
in the same manner it may be demonstrated, that the straight lines

DC, DD are each of them equal to DA or DB :
therefore the four straight lines DA, DB, DC, DD are equal to one

another ;
and the circle described from the center D, at the distance of one
of them, will pass through the extremities of the other three, and be
described about the square ABCD. q.e.f.

PROPOSITION X. PROBLEM.

To describe an isosceles triangle, having each of the angles at the base
double of the third angle.

Take any straight line AB, and divide it in the point C, (ll. 11.)
so that the rectangle AB, BC may be equal to the square on CA ;
and from the center A, at the distance AB, describe the circle BDD,
in which place the straight line BD equal to A C, which is not greater
than the diameter of the circle BDD; (iv. 1.)
and join DA.
Then the triangle ABD shall be such as is required,
that is, each of the angles ABD, ADB shall be double of the angle
Join DC, and about the triangle ^D (7 describe the circle A CD. (iv. 5.)
And because the rectangle AB,BC is equal to the square on AC,
and that ^ C is equal to BD, (constr.)
the rectangle AB, BC is equal to the square on J3D: (ax. 1.)
and because from the point B, without the circle A CD, two straight
lines BCA, BD are drawn to the circumference, one of which cuts, and

BOOK IV. PROP. X, XI. 183

the other meets the circle, and that the rectangle AB, BC, contained
by the whole of the cutting line, and the part of it without the circle,
is equal to the square on BD which meets it ;

therefore the straight line BD touches the circle ^ CD: (ill. 37.)
and because BD touches the circle, and DC is drawn from the

point of contact D,
the angle BD C is equal to the angle DA C in the alternate segment
of the circle: (ill. 32.)

to each of these add the angle CD A ;
therefore the whole angle BDA is equal to the two angles CD A,
DAC: (ax. 2.)
but the exterior angle J5 CD is equal to the angles CD A, DA C; (l. 32.)
therefore also BDA is equal to BCD : (ax. 1.)
but BDA is equal to the angle CBD, (i. 5.)
because the side AD is equal to the side AB ;
therefore CBD, or DBA, is equal to BCD; (ax. 1.)
and consequently the three angles BDA, DBA, BCD are equal to
one another :

and because the angle DBCis equal to the angle BCD,
the side BD is equal to the side DC: (l. 6.)
but BD was made equal to CA ;
therefore also CA is equal to CD, (ax. 1.)
and the angle CD A equal to the angle DAC', (l. 5.)
therefore the angles CD A, D^ C together, are double of the angle
DAC:
but BCD is equal to the angles CD A, DAC; (l. 32.)
therefore also BCD is double o^ DAC:
and BCD was proved to be equal to each of the angles BDA, DBA ;
therefore each of the angles BDA, DBA is double of the angle DAB,
Wherefore an isosceles triangle ABD has been described, having
each of the angles at the base double of the third angle. Q.e.f.

PROPOSITION XI. PROBLEM.

To inscribe an equilateral and equiangular pentagon in a given circle.

Let ABCDE be the given circle.
It is required to inscribe an equilateral and equiangular pentagon
in the circle ABCDE.

Describe an isosceles triangle FGH, having each 'of the angles at
G, ^double of the angle at F; (iv. 10.)

and in the circle ABCDE inscribe the triangle ACD equiangular
to the triangle FGH, (iv. 2.)

so that the angle CAD may be equal to the angle at F,
and each of the angles A CD, CD A equal to the angle at G or H\

1S4 Euclid's elements.

wherefore each of the angles ^ CD, CD A is double of the angle CAD.

Bisect the angles ^ CD, CD A by the straight lines CE, DB; (i. 9.)

and join AB, BC, DE, EA.

G H

Then ABCDE shall be the pentagon required.
Because each of the angles A CD, CD A is double of CAD,
and that they are bisected by the straight lines CE, DB ;
therefore the five angles DAC, ACE, ECD, CDB, BDA are

equal to one anotlier :
but equal angles stand upon equal circumferences ; (ill. 26.)
therefore the five circumferences AB,BC, CD, DE, EA are equal
to one another :
and equal circumferences are subtended by equal straight lines ; (ill. 29.)
therefore the five straight lines AB, BC, CD, DE, EA are equal
to one another.

Wherefore the pentagon ABCDE is equilateral.

It is also equiangular :

for, because the circumference AB h equal to the circumference DE,

if to each be added BCD,

the whole A BCD is equal to the whole EDCB: (ax. 2.)

but the angle AED stands on the circumference ABCD ;

and the angle BAE on the circumference EDCB ;

therefore the angle BAE is equal to the angle AED : (ill. 27.)

ibr the same reason, each of the angles ABC, BCD, CDE is equal

to the angle BAE, or AED :

therefore the pentagon ABCDE is equiangular ;
and it has been shewn that it is equilateral :
wherefore, in the given circle, an equilateral and equiangular pentagon
has been described, q.e.f.

PROPOSITION XII. PROBLEM.

To describe an equilateral and equiangular pentagoti about a given circle.

Let ABCDE be the given ci;-cle.
It is required to describe an equilateral and equiangular pentagon

Let the angular points of a pentagon, inscribed in the circle, by the
last proposition, be in the points A, B, C, D, E,
so that the circumferences AB,BC, CD, DE, EA are equal ; (iv. 11.)
and through the points A, B, C, D, E draw GET, HK, XL, LM,
MG toucliing the circle; (ill. 17.)
the figure GHKL3I ^]\?l\\ be the pentagon required.
Take the center F, and join FB, FK, EC, FL, FD.
And because the straight line KL touches the circle ABCDE in
the point C, to which EC is drawn from the center F,
EC is perpendicular to XL, (ill. 18.)

BOOK IV. PROP. XII.

185

therefore each of the angles at C is a right angle :
the same reason, the angles at the points B^ D are right angles :
G

K C L

and because FCK is a right angle,
the square on FKh equal to the squares on FC, CX\ (l. 47.)
for the same reason, the square on FK is equal to the squares on

FB, BK:
therefore the squares on jFC, CK are equal to the squares on FB^
BK; (ax. 1.)
of which the square on FC is equal to the square on FB ;
therefore the remaining square on CK is equal to the remaining square
on BK, (ax. 3.) and the straight line CS" equal to BK:
and because FB is equal to FC, and FK common to the triangles

BFK, CFK,

the two BF, FK are equal to the two CF, FK, each to each :

and the base BK was proved equal to the base KC:

therefore the angle BFK is equal to the angle KFC, (i. 8.)

and the angle ^^i^^to FKC: (i. 4.)

wherefore the angle BFC is double of the angle KFC,

and BKC double of FKCi

for the same reason, the angle CFD is double of the angle CFL,

and CLD double of CLF-.

and because the circumference BC is equal to the circumference CD,

the angle BFC is equal to the angle CFI>; (ill. 27.)

and BFC is double of the angle KFC,

and CFD double of CFL-,

therefore the angle KFC is equal to the angle CFL: (ax. 7.)

and the right angle FCK is equal to the right angle FCL ;

therefore, in the two triangles FKC, FLC, there are two angles of the

one equal to two angles of the other, each to each ;
and the side FC which is adjacent to the equal angles in each, is com-
mon to both ;
therefore the other sides are equal to the other sides, and the third

angle to the third angle : (I. 26.)
therefore the straight line KC is equal to CL, and the angle FKC
to the angle FLC-.

and because KC is equal to CL,

KL is douhle o{ KC

In the same manner it may be shewn that HK is double of BK :

and because BK is equal to KC, as was demonstrated,

and that KL is double of KC, and BK double of BK,

therefore UK is equal to KL : (ax. 6.)

in like manner it may be shewn that GH, GM, ML are each of them

equal to HK, or KL :

186 EUCLID'S ELEMENTS.

therefore the pentagon GHKLM is equilateral.

It is also equiangular :

for, since the angle FKC is equal to the angle FZC,

and that the angle HKL is double of the angle FKC,

and jfirXil[f double of FZC, as was before demonstrated;

therefore the angle JiKL is equal to KLM-. (ax. 6.)

and in like manner it may be shewn,

that each of the angles KHG, UGH, GML is equal to the angle

HKL or KLM'.
therefore the five angles GHK, HKL, KLM, LMG, MGH heing
equal to one another,

the pentagon GHKL3I is equiangular :

and it is equilateral, as was demonstrated;

and it is described about the circle ABCDE. Q.E.F.

PROPOSITION XIII. PROBLEM.
To inscribe a circle in a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon.
It is required to inscribe a circle in the pentagon ABCDE.

Bisect the angles BCD, CDE by the straight lines CF, DF, (l. 9.)
and from the point F, in which they meet, draw the straight lines FB,
FA,FE:

therefore since ^Cis equal to CD, (hj'p.)

and CF common to the triangles BCF, DCF,

the two sides BC, CF are equal to the two DC, CF, each to each ;

and the angle BCFU equal to the angle DCF; (constr.)

therefore the base BF is equal to the base FD, (i. 4.)

and the other angles to the other angles, to which the equal sides are

opposite :

therefore the angle CBF is equal to the angle CDF:

and because the angle CDE is double of CDF,
and that CDE is equal to CBA, and CDF to CBF;

CBA is also double of the angle CBF;

therefore the angle ABE is equal to the angle CBF;

wherefore the angle ABC is bisected by the straight line BF:

in the same manner it may be demonstrated,

that the angles BAE, A ED, are bisected by the straight lines AF, FE.

From the point F, draw EG, FH, FK, EL, Filf perpendiculars to

the straight lines AB, BC, CD, DE, EA : (i. 12.)

and because the angle HCF is equal to KCF, and the right angle
FHC equal to the right angle FKC)

I

BOOK IV. PROP. XllI, XIV. 187

therefore in the triangles FHC, FKC, there are two angles of the one

equal to two angles of the other, each to each ;
and the side FC, which is opposite to one of the equal angles in each,
is common to both ;
therefore the other sides are equal, each to each ; (l. 26.)
wherefore the perpendicular FHis equal to the perpendicular FK:
in the same manner it may be demonstrated, that FL, FM, FG are
each of them equal to FH, or FK\

therefore the five straight lines FG^ FH, FK, FL, FM are equal
to one another :
wherefore the circle described from the center F, at the distance of
one of these five, will pass through the extremities of the other four,
and touch the straight lines AB, BC, CD, DF, FA,

because the angles at the points G, H, K, L, M are right angles,
and that a straight line drawn from the extremity of the diameter of
a circle at right angles to it, touches the circle ; (ill. 16.)
therefore each of the straight lines AB, BC, CD, DF, FA touches
the circle :
wherefore it is inscribed in the pentagon ABCDE. Q.e.f.

PROPOSITION XIV. PROBLEM.

To describe a circle about a given equilateral and equiangular pentagon.

Let ABCDFhe the given equilateral and equiangular pentagon.
It is required to describe a circle about ABCDF.

A

Bisect the angles BCD, CDF by the straight lines CF, FD, (l. 9.)
and from the point F, in which they meet, di-aw the straight lines FB,
FA, FE, to the points B, A, F.
It may be demonstrated, in the same manner as the preceding pro-
position,

that the angles CBA, BAF, AED are bisected by the straight lines
FB, FA, FE.

And because the angle BCD is equal to the angle CDE,

and that FCD is the half of the angle BCD,

and CDF the half of CDE;

therefore the angle FCD is equal to FDC\ (ax. 7.)

wherefore the side CF is equal to the side FD : (l. 6.)

in like manner it may be demonstrated,

that FB, FA, FE, are each of them equal to EC or FD :

therefore the five straight lines FA, FB, FC, FD, FE, are equal to

one another ;
and the circle described from the center i^, at the distance of one of
them, will pass through the extremities of the other four, and be de-
scribed about the equilateral and equiangular pentagon ABCDE.

Q.E.F.

188 Euclid's elements.

PROPOSITION XV. PROBLEM.
To inscribe an equilateral and equianffular hexago7i in a given circle.

Let AJBCDJEFhe the given circle.
It is required to inscribe an equilateral and equiangular hexagon in it.

A
^" ,B

Find the center G of the circle ABCDEF,

and draw the diameter AGD; (in. 1.)

and from D, as a center, at the distance Z)(r, describe the circle EGCHj

join EG, CG, and produce them to the points B, F;

and join AB, BC\ CD, BE, EF, FA :

the hexagon AB CDEF shall be equilateral and equiangular.

Because G is the center of the circle ABCDEF,

GE is equal to GD :

and because D is the center of the circle EGCS,

DEis equal to DG:

wherefore GE is equal to ED, (ax. 1.)

and the triangle EGD is equilateral ;

and therefore its three angles EGD, GDE, DEG, are equal to one

another: (i. 5. Cor.)

but the three angles of a triangle are equal to two right angles ; (l. 32.)

therefore the angle EGD is the third part of two right angles :

in the same manner it may be demonstrated,
that the angle DGC is also the third part of two right angles :
and because the straight line (?C makes with EB the adjacent angles
EGC, CGB equal to two right angles ; (i. 13.)
the remaining angle CGB is the third part of two right angles:
therefore the angles EGD, DGC, CGB are equal to one another:
and to these are equal the vertical opposite angles BGA, A GF, FGE :
(I. 15.)
therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE,
are equal to one another :
but equal angles stand upon equal circumferences ; (in. 26.)
therefore the six circumferences AB, BC, CD, DE, EF, FA are equal
to one another :
and equal circumferences are subtended by equal straight lines :
(III. 29.)

therefore the six straight lines are equal to one another,
and the hexagon ABCDEF h equilateral.
It is also equiangular:
for, since the circumference AF is equal to ED,
to each of these equals add the circumference ABCD ;
therefore the whole cu-cumference FAB CD is equal to the whole
EDCBA'.

BOOK IV. PROP. XV, XVI. 189

and the angle FED stands upon the circumference FABCDy
and the angle AFE upon EDCBA ;
therefore the angle AFE is equal to FED : (lii. 27.)
in the same manner it may be demonstrated,
that the other angles of the hexagon ABCDEF are each of them
equal to the angle AFE or FED : therefore the hexagon is equi-
angular ; and it is equilateral, as was shewn ;
and it is inscribed in the given circle ABCDEF. q.e.f.
Cor. â€” From this it is manifest, that the side of the hexagon is
equal to the straight line from the center, that is, to the semi-diameter
of the circle.

And if through the points A,B, C, D, E, F there be drawn straight
lines touching the circle, an equilateral and equiangular hexagon will
be described about it, which may be dembnstrated from what has been
said of the pentagon: and likewise a circle may be inscribed in a given
equilateral and equiangular hexagon, and circumscribed about it, by a
method like to that used for the pentagon.

PROPOSITION XYI. PROBLEM.

To inscribe an equilateral and equiangular quindecagon in a given circle.

Let A BCD be the given circle.
It is required to inscribe an equilateral and equiangular quindeca-
gon in the circle ABCD.

Let A Che the side of an equilateral triangle inscribed in the circle, (l v. 2.)
and AB the side of an equilateral and equiangular pentagon inscribed
in the same; (iV. 11.)
therefore, of such equal parts as the whole circumference ABCDF
contains fifteen,
the circumference ABC, being the third part of the whole, contains five ;
and the circumference AB, which is the fifth part of the whole, con-
tains three ;

therefore BC, their difierence, contains two of the same parts:
bisect ^ C in ^; (iii. 30.)
therefore BE, EC are, each of them, the fifteenth part of the whole
circumference ABCD :
therefore if the straight lines BE, EC be drawn, and straight lines
equal to them be placed round in the whole circle, (iv. 1.) an equi-
lateral and equiangular quindecagon will be inscribed in it. Q. E. F.

And in the same manner as was done in the pentagon, if through
the points of division made by inscribing the quindecagon, straight
lines be drawn touching the circle, an equilateral and equiangular
quindecagon will be described about it : and likewise, as in the pen-
tagon, a circle may be inscribed in a given equilateral and equiangular

NOTES TO BOOK IV.

The Fourth Book of the Elements contains some particular cases of
four general problems on the inscription and the circumscription of tri-
angles and regular figures in and about circles. Euclid has not given
any instance of the inscription or circumscription of rectilineal -figures
in and about other rectilineal figures.

Any rectilineal figure, of five sides and angles, is called a pentagon ;
of seven sides and angles, a heptagon ; of eight sides and angles, an octa-
gon ; of nine sides and angles, a nonagon ; of ten sides and angles, a
decagon ; of eleven sides and angles, an undecagon ; of twelve sides and
angles, a duodecagon ; of fifteen sides and angles, a quindecagon, &c.

These figures are included under the general name of polygons ; and
are called equilateral, when their sides are equal ; and equiangular, when
their angles are equal ; also when both their sides and angles are equal,
they are called regular polygom.

Prop. III. An objection has been raised to the construction of this
problem. It is said that in this and other instances of a similar kind,
the lines which touch the circle at A, B, and C, should be proved to meet
one another. This may be done by joining /IB, and then since the angles
KAM, KB M axe equal to two right angles (in. 18.), therefore the angles
BAM, ABM are less than two right angles, and consequently (ax. 12.),
AM and J53/must meet one another, when produced far enough. Similarly,
it may be shewn that AL and CL, as also CN and By meet one another.

Prop. V. is the same as "To describe a circle passing through three
given points, provided that they are not in the same straight line."

The corollary to this proposition appears to have been already de-
monstrated in Prop. 31. Book in.

It is obvious that the square described about a circle is equal to
double the square inscribed in the same circle. Also that the circum-
scribed square is equal to the square on the diameter, or four times the
square on the radius of the circle.

Prop. VII. It is manifest that a square is the only right-angled paral-
lelogram which can be circumscribed about a circle, but that both a
rectangle and a square may be inscribed in a circle.

Prop. X. By means of this proposition, a right angle may be divided
into five equal parts.

and synthesis, and that all Euclid's direct demonstrations are synthetic,
properly so called. There is however a single exception in Prop. 16.
Book IV, where the analysis only is given of the Problem. The two
methods are so connected in all processes of reasoning, that it is very
difficult to separate one from the other, and to assert that this process is
really sxjnthetic, and that is really analytic. In every operation performed
in the construction of a problem, there must be in the mind a knowledge
of some properties of the figure which suggest the steps to be taken in
the construction of it. Let any Problem be selected from Euclid, and at
each step of the operation, let the question be asked, " Why that step
is taken r" It will be found that it is because of some known property
of the required figure. As an example will make the subject more clear
to the learner, the Analysis of Euc. iv. 10, is taken from the ''Analysis of
Problems'* in the larger edition of the Euclid, and to which the learner
is referred for more complete information.

In Euc. IV. 10, there are five operations specified in the construc-
tion : â€”

(1) Take any straight line ^4i?,

NOTES TO BOOK IV. 191

(2) Divide the line AB in C, so that the rectangle AB, BC, may be
equal to the square on AC.

(3) Describe the circle BDE with center J and radius AB.

(4) Place the line BD in that circle, equal to the line AC.

(5) Join the points A, D,

Why should either of these operations be performed rather than any
others ? And what will enable us to forsee that the result of them will
be such a triangle as was required ? The demonstration affixed to it by
Euclid does undoubtedly prove that these operations must, in conjunction,
produce such a triangle : but we are furnished in the Elements with no
obvious reason for the adoption of these steps, unless we suppose them
accidental. To suppose that all the constructions, even the simpler ones,
are the result of accident only, would be supposing more than could be
shewn to be admissible. No construction of the problem could have
been devised without a previous knowledge of some of the properties of
the figure. In fact, in directing the figure to be constructed, we assume
the possibility of its existence ; and we study the properties of such a
figure on the hypothesis of its actual existence. It is this study of the
properties of the figure that constitutes the Analysis of the problem.

Let then the existence of a triangle BAD be admitted, which has each
of the angles ABD, ADB double of the angle BAD, in order to ascertain
any properties it may possess which would assist in the construction of
such a triangle.

Then, since the angle ADB is double of BAD^ if we draw a line DC
to bisect ADB and meet AB in C, the angle ADC will be equal to CAD ;
and hence (Euc. i. 6.) the sides AC, CD are equal to one another.

Again, since we have three points A, C, Z), not in the same straight
line, let us examine the efiect of describing a circle through them : that
is, describe the circle ACD about the triangle A CD. (Euc. iv. 5.)

Then, since the angle ADB has been bisected by DC, and since ADB
is double of DJB, the angle CDB is equal to the angle DAC in the alter-
nate segment of the circle ; the line BD therefore coincides with a tangent
to the circle at D. (Converse of Euc. in. 32.)

Whence it follows, that the rectangle contained by AB, BC, is equal
to the square on BD. (Euc. iii. 36.)

But the angle BCD is equal to the two interior opposite angles CAD,
CD A; or since these are equal to each another, BCD is the double of
C/^/), that is, of BAD. And since ABD is also double of BAD, by the
conditions of the triangle, the angles BCD, CBD are equal, and BD is
equal to DC, that is, to AC.

It has been proved that the rectangle AB, BC, is equal to the square
on BD ; and hence the point C in AB, found by the intersection of the
bisecting line DC, is such, that the rectangle JB, BC is equal to the
square on AC. (Euc. ii. II.)

Finally, since the triangle ABD is isosceles, having each of the angles
ABD, ADB double of the same angle, the sides AB, AD are equal, and
hence the points B, D, are in the circumference of the circle described
about A with the radius AB. And since the magnitude of the triangle
is not specified, the line AB may be of any length w hatever.

From this ** Analysis of the problem," which obviously is nothing
more than an examination of the properties of such a figure supposed to
exist already, it will be at once apparent, why those steps which are
prescribed by Euclid for its construction, were adopted.

The line AB is taken of any length, because the problem does not
prescribe any specific magnitude to any of the sides of the triangle.

19^ Euclid's elements.

The circle BDE is described about A with the distance AB, because
the triangle is to be isosceles, having AB for one side, and therefore the
other extremity of the base is in the circumference of that circle.

The line AB is divided in C, so that the rectangle AB^ BC shall be
equal to the square on ACy because the base of the triangle must be equal
to the segment AC.

And the line AD is drawn, because it completes the triangle, two of
whose sides, ABy BI) are already drawn. l

Whenever we have reduced the construction to depend upon problems â–
which have been already constructed, our analysis may be terminated ; *
as was the case where, in the preceding example, we arrived at the
division of the line AB in C ; this problem having been already con-
structed as the eleventh of the second book.

Prop. XVI. The arc subtending a side of the quindecagon, may be