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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Let ABC he the given segment of a circle.

It is required to describe the circle of which it is the segment.

Bisect AC in D, (i. 10.) and from the point D draw DB at right
angles to AC, (i. 11.) and join AB.

First, let the angles ABD, BAD be equal to one another :

ÂŁP\

then the straight line DA is equal to DB, (l. 6.) and therefore, to DC;
and because the three straight lines DA, DB, DC are all equal,

therefore D is the center of the circle, (ill. 9.)
From the center D, at the distance of any of the three DA, DB,
DC, describe a circle;

this shall pass through the other points ;
and the circle of which ABC is a segment has been described :

I

BOOK 111. PROP. XXVI. 141

,iid because the center D is in AC, the segment ABCh a semicii'cle.
But if the angles ABD, BAD are not equal to one another:

^0

IE

D C

I

W^m at the point A, in the straight line AB,

^" make the angle BAE equal to the angle ABD, (r. 23.)

and produce BD, if necessary, to meet AE in E, and join EC.
Because the angle ABE is equal to the angle BAE,
therefore the straight line EA is equal to EB : (i. 6.)
and because AD is equal to DCy and DE common to the triangles

the two sides AD, DE, are equal to the two CD, DE, each to each ;

and the angle ADE is equal to the angle CDE.

for each of them is a right angle ; (constr.)

therefore the base EA is equal to the base EC-, (i. 4.)

but EA was shewn to be equal to EB :

wherefore also EB is equal to EC: (ax. 1.)

and therefore the three straight lines EA, EB, EC are equal to one

another :

wherefore E is the center of the circle, (ill. 9.)
From the center E, at the distance of any of the three EA, EB,

EC, describe a circle ;

this shall pass through the other points ;
and the circle of which ABC is a segment, is described.
And it is evident, that if the angle ABD be greater than the angle
BAD, the center E falls without the segment ABC, which therefore
is less than a semicircle :

but if the angle ABD be less than BAD, the center E falls within
the segment ABC, which is therefore greater than a semicircle.

Wherefore a segment of a circle being given, the circle is described
of which it is a segment. Q. e. f.

PROPOSITION XXVI. THEOREM.

In equal circles, equal angles stand upon equal arcs, whether the angles be
at the centers or circumferences.

Let ABC, DEFhe equal circles,

and let the angles BGC, EIIF at their centers,

and BA C, EDF at their circumferences be equal to each other.

Then the arc BKC shall be equal to the arc ELF.

A D

14^ Euclid's elements.

Join BC, EF.

And because the circles ABC, DBF are equal,

the straight lines drawn from their centers are equal : (ill. def. 1.)

therefore the two sides BG, GC, are equal to the two JEJff MF, each

to each :

and the angle at G is equal to the angle at H; (hyp.)

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

And because the angle at A is equal to the angle at B>, (hyp.)

the segment BACis similar to the segment BBF: (ill. def. 11.)

and they are upon equal straight lines BC, EF:

but similar segments of circles upon equal straight lines, are equal to

one another, (iii. 24.)

therefore the segment BA C is equal to the segment EDF:
but the whole circle ABC is equal to the whole DEF-, (hyp.)
therefore the remaining segment BKC is equal to the remaining seg-
ment ELF, (i. ax. 3.)

and the arc BKC to the arc ELF.
Wherefore, in equal circles, &c. q.e.d.

PEOPOSITION XXVII. THEOREM.

In equal circles, the angles which stand upon equal arcs, are equal to one
another, whether they he at the centers or circumferences.

Let ABC, DEF he equal circles,

and let the angles BGC, EHF at their centers,

and the angles BA C, EDF at their circumferences,

stand upon the equal arcs BC, EF.

Then the angle J?G^Cshall be equal to the angle EHF,

and the angle BA C to the angle EDF.

If the angle BGC he equal to the angle ERF,
it is manifest that the angle BA C is also equal to EDF. (ill. 20. and
I. ax. 7.)

But, if not, one of them must be greater than the other :

if possible, let the angle BGC he greater than EHF,

and at the point G, in the straight line BG,

make the angle BGK equal to the angle EHF. (I. 23.)

Then because the angle BGX is equal to the angle EHF,

and that equal angles stand upon equal arcs, when they are at the

centers ; (ill. 26.)

therefore the arc BK is equal to the arc EF:

but the arc EF is equal to the arc BC; (hyp.)

therefore also the arc BX is equal to the arc BC,

the less equal to the greater, which is impossible: (l. ax. 1.)

BOOK 111. PROP. XXVIII, XXIX.

143

therefore the angle JBGCis not unequal to the angle EHF;

that is, it is equal to it :

but the ano-le at A is half of the angle BGC, (ill. 20.)

and the angle at D, half of the angle EHF-,

therefore the angle at A is equal to the angle at D. (l. ax. 7.)

Wherefore, in equal circles, &c. Q. E. D.

PROPOSITION XXVIII. THEOREM.

In equal circles, equal straight lines cut off equal arcs, the greater equal
to the greater, and the less to the less.

Let ABC, BUFhe equal circles,
and BC, EF equal straight lines in them, which cut off the two greater
arcs BAC, EDF, and the two less BGC, EHF.
Then the greater arc BA C shall be equal to the greater EDF^
and the less arc BGC to the less EHF.

Take K,L, the centers of the circles, (iii. 1.) and join BK, KC, EL, LF.

Because the circles ABC, DEFdiYe equal,

the straight lines from their centers are equal : (iii. def. 1.)

therefore BK, KC are equal to EL, LF, each to each :

and the base BC is equal to the base EF, in the triangles BCK, EEL-,

therefore the angle BKCh equal to the angle ELF: (i. 8.)

but equal angles stand upon equal arcs, when they are at the

centers : (lli. 26.)

therefore the arc BGCis equal to the arc EHF:

but the whole circumference ABCis equal to the whole EDF', (hyp.)

therefore the remaining part of the circumference,

viz. the arc BA C, is equal to the remaining part EDF. (l. ax. 3.)

Therefore, in equal circles, &c. q.e.d.

PROPOSITION XXIX. THEOREM.

circles, equal arcs are subtended by equal straight lines.

Let ABC, DEF he equal circles,

and let the arcs BGC, EHF also be equal,

and joined by the straight lines BC, EF.

Then the straight line BC shall be equal to the straight line EF.

A D

^ K \ / L

144 Euclid's elements.

Take K, L, (ill. 1.) the centers of the circles, and join BK,KC,EL, LI.

Because the arc BGCis equal to the arc JEJIF,

therefore the angle BKCis equal to the angle ULF: (III. 27.)

and because the circles ABC, DJEF, are equal,

the straight lines from their centers are equal; (ill. def. 1.)

therefore BK, KC, are equal to EL, LF, each to each :

and they contain equal angles in the triangles BCK, FFL;

therefore the base BCis equal to the base FF. (I. 4.)

Therefore, in equal circles, &c. Q.E. D.

PROPOSITION XXX. PROBLEM.

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

Let ADB be the given arc :
it is required to bisect it.

A C B

Join AB, and bisect it in C; (I. 10.)

from the point C draw CD at right angles to AB. (l. 11.)

Then the arc ADB shall be bisected in the point D.

And because ^ C is equal to CB,

and CD common to the 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 the angle BCD,

because each of them is a right angle :

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

But equal straight lines cut off equal arcs, (ill. 28.)

the greater arc equal to the greater, and the less arc to the less ;

and the arcs AD, DB are each of them less than a semicircle ;

because DC, if produced, passes through the center: (ill. 1. Cor.)

therefore the arc AD is equal to the arc DB.

Therefore the given arc ADB is bisected in D. q.e.f.

PROPOSITION XXXI. THEOREM.

In a circle, the angle in a semicircle is a right angle ; hut the angle in a
segment greater than a semicircle is less than a right angle ; and the angle
in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is ^C, and center E,

and let CA be drawn, dividing the circle into the segments ABC, ADC.

Then the angle in the semicircle BA C shall be a right angle ;

and the angle in the segment ABC, which is greater than a semicircle,

shall be less than a right angle ;
and the angle in the segment ADC, which is less than a semicircle,
shall be greater than a right angle.

BOOK 111. PRor. XXXI, XXXII. 145

/

â– ^D

Join AE, and produce BA to F.
First, because JEB is equal to UA, (l. def. 15.)
the angle UAJB is equal to UJBA ; (i. 5.)
also, because UA is equal to EC,
the angle ECA is equal to EAC]
â€˘wherefore the whole angle ÂŁ AC is equal to the two angles EBA,
ECA ; (I. ax. 2.)
but FAC, the exterior angle of the triangle ABC, is equal to the two
angles EBA, ECA ; (l. 32.)

therefore the angle BACis equal to the angle FAC; (ax. 1.)
and therefore each of them is a right angle : (l. def. 10.)
wherefore the angle BA C in a semicircle is a right angle.
Secondly, because the two angles ABC, BAC of the triangle
ABC are together less than two right angles, (l. 17.)

and that ^^Chas been proved to be a right angle;
therefore ^_BCmust be less than a right angle :
and therefore the angle in a segment ABC greater than a semicircle,
is less than a right angle.
And lastly, because A BCD is a quadrilateral figure in a circle,
any two of its opposite angles are equal to two right angles : (ill. 22.)
therefore the angles ABC, ADC, are equal to two right angles :
and ^jBChas been proved to be less than a right angle;
wherefore the other ADC is greater than a right an^e.
erefore, in a circle the angle in a semicircle is a right angle ; &c. Q.E.D.
Cor. From this it is manifest, that if one angle of a triangle be
equal to the other two, it is a right angle : because the angle adjacent
to it is equal to the same two ; (l. 32.) and when the adjacent angles
are equal, they are right angles. (l. def. 10.)

PROPOSITION XXXII. THEOREM.

If a straight line touch a circle, and from the point of contact a straight
line be draicn meeting the circle ; the angles tchich this line makes icith the
line touching the circle shall be equal to the angles which are in the alter-
nate segments of the circle.

Let the straight line JE'i^ touch the circle ABCD in B,
and from the point B let the straight line BD be drawn, meeting
the cii'cumference in D, and dividing it into the segments DCi?, i>^i/,
of which DCB is less than, and DAB greater than a semcircle.

Then the angles which BD makes with the touching line EF,
shall be equal to the angles in the alternate segments of the circle ;
that is, the angle DBF shall be equal to the angle which is in the
segment DAB,

H

1

146

]

segment BCB.

and the angle DBE shall be equal to the angle in the alternate
DP"

From the point B draw BA at right angles to EF, (l. 11.) meeting
the circumference in A ;

take any point Cin the arc DB, and join AD, DC, CB.
Because the straight line EF touches the circle A BCD in the
point B,

and BA is drawn at right angles to the touching line from the
point of contact B,

the center of the circle is in BA : (ill. 19.)
therefore the angle ADB in a semidrcle is a right angle: (ill. 31.)
and consequently the other two angles BAD, ABD, are equal to
a right angle ; (i. 32.)

but ABF is likewise a right angle ; (constr.)
therefore the angle ABF is equal to the angles BAD, ABD: (l.ax. 1.)

take from these equals the common angle ABD:

therefore the remaining angle DBFis equal to the angle BAD, (l. ax. 3.)

which is in BDA, the alternate segment of the circle.

And because ABCD is a quadrilateral figure in a circle,

the opposite angles J?^Z>,jB CD are equal to two right angles: (III. 22.)

but the angles DBF, DBE are likewise equal to two right angles ;

(I. 13.)
therefore the angles DBF, DBE are equal to the angles BAD,
BCD, (I. ax. 1.)

and DBF has been proved equal to BAD ;
therefore the remaining angle DBE is equal to the angle BCD in
BDC, the alternate segment of the circle. (l. ax. 2.)
Wherefore, if a straight line, &c. Q. E. D.

PKOPOSITION XXXIII. PROBLEM.

Upon a given straight line to describe a segment of a circle, which shall
contain an angle equal to a given rectilineal angle.

Let ABhe the given straight line,
and the angle C the given rectilineal angle.
It is required to describe upon the given straight line ^5, a segment
of a circle, which shall contain an angle equal to the angle C
First, let the angle C be a right angle.

XX

{

BOOK III. PROP. XXXllI, XXXIV. 147

Bisect AB in F, (l. 10.)
and from the center i^, at the distance i*^^, describe the semicircle ^J^JB,
and draw AH, BIl to any point II in the circumference.

Therefore the angle AHB in a semicircle is equal to the right
angle C. (ill. 31.)

But if the angle Cbe not a right angle :

at the point A, in the straight line AB,
make the angle BAD equal to the angle C, (i. 23.)
and from the point A di-aw AB at right angles to AD\ (l. 1 1.)
bisect AB in F, (i. 10.)
and from i^draw FG at right angles to AB, (l. 11.) and join GB.
Because AF is equal to FB, and FG common to the triangles
AFG, BFG,
the two sides AF, FG are equal to the two BF, FG, each to each,
and the angle AFG is equal to the angle BFG; (i. def. 10.)

therefore the base AG\^ equal to the base GB', (i. 4.)
and the circle described from the center G, at the distance GA,
shall pass through the point B :

let this be the circle AHB.
The segment AHB shall contain an angle equal to the given rec-
tilineal angle C.

Because from the point A the extremity of the diameter AE,
AD is drawn at right angles to AE,

therefore AD touches the circle: (iii. 16. Cor.)
and because AB, drawn from the point of contact A, cuts the circle,
the angle DAB is equal to the angle in the alternate segment
AHB: (111.32.)

but the angle DAB is equal to the angle (7; (constr.)
therefore the angle C is equal to the angle in the segment AHB.
Wherefore, upon the given straight line AB, the segment AHB
of a circle is described, which contains an angle equal to the given
angle C. Q.e.f,

PROPOSITION XXXIV. PROBLEM.

From a given circle to cut off a segment, which shall cortiain an angle
equal to a given rectilineal angle.

Let ABC he the given circle, and D the given rectilineal angle.
It is required to cut off from the circle ABC a segment that shall
contain an angle equal to the given angle D.

h2

â–

148 Euclid's elements.

Draw the straight line EF touching the circle ABC in any point B,
(III. 17.)

and at the point B, in the straight line BF,
make the angle i^^C equal to the angle D. (l. 23.)
Then the segment BA C shall contain an angle equal to the given
angle D.

Because the straight line ^.F touches the circle ABC,
and BCh drawn from the point of contact J5,
therefore the angle FBC is equal to the angle in the alternate
segment BA C of the circle : (IIL 32.)
but the angle FBCis equal to the angle D ; (constr.)
therefore the angle in the segment BA C is equal to the angle

D. (I. ax. 1.)
Wherefore from the given circle ABC, the segment BAC is cut
off, containing an angle equal to the given angle D. Q. E. F.

PROPOSITION XXXV. THEOREM.

If two straight lines cut one another within a circle, the rectangle contained
by the segments of one of them, is equal to the rectangle contained by the
segments of the other.

Let the two straight lines A C, BD, cut one another in the point
E, within the circle A BCD.

Then the rectangle contained by AE, EC shall be equal to the
rectangle contained by BE, ED.

First, if A C, BD pass each of them through the center, so that E
is the center ;

it is evident that since AE, EC BE, ED, being all equal, (I. def. 15.)
therefore the rectangle AE, ECis equal to the rectangle BE, ED.

Secondly, let one of them BD pass through the center, and cut the
other A C, which does not pass through the center, at right angles, in
the point E.

D

BOOK III. PROP. XXXV. 149

Then, if BD be bisected in F,

jPIs the center of the circle ABCD.

Join AF.

Because BD which passes through the center, cuts the straight
line A C, which does not pass through the center, at right angles in F,
therefore AE is equal to EC: (in. 3.)
and because the straight line BD is cut into two equal parts in the

point F, and into two unequal parts in the point E,
therefore the rectangle BE, ED, together with the square on EF,
is equal to the square on FB ; (ll. 5.)

that is, to the square on FA :
but the squares on AE, EF, are equal to the square on FA : (l. 47.)
therefore the rectangle BE, ED, together with the square on EF,
is equal to the squares on AE, EF: (l. ax. 1.)
take away the common square on EF,
and the remaining rectangle BE, ED is equal to the remaining
square on AE; (l. ax. 3.)

that is, to the rectangle AE, EC.
Thirdly, let BD, which passes through the center, cut the other A C,
which does not pass through the center, in E, but not at right angles.

Then, as before, if BD be bisected in F,

F is the center of the circle.

Join AF, and from JPdraw EG perpendicular to AC; (l. 12.)

therefore ^6^ is equal to GC; (III. 3.)
wherefore the rectangle AE, EC, together with the square on EG\
is equal to the square on AG: (ii. 5.)

to each of these equals add the square on GF;

therefore the rectangle AE, EC, together with the squares on EG,

GF, is equal to the squares on A G, GF; (i. ax. 2.)

but the squares on EG, GF, are equal to the square on EF; (i. 47.)

and the squares on AG, GF Sire equal to the square on AF:

therefore the rectangle AE, EC, together with the square on EF,

is equal to the square on ^i^;

that is, to the square on FB :
but the square on FB is equal to the rectangle BE, ED, together
with the square on EF; (ii. 5.)
therefore the rectangle AE, EC, together with the square on EF,
is equal to the rectangle BE, ED, together with the square on
EF; (I. ax. 1.)

take away the common squai'e on EF,
and the remaining rectangle AE, EC, is therefore equal to the re-
maining rectangle BE, ED. (ax. 3.)
Lastly, let neither of the straight lines A C, BD pass through the
center.

150

EUCLID S ELEMENTS.

Take the center F, (ill. 1.)
and through Tl the intersection of the straight lines -4 C, DJ?,

draw the diameter GEFH.
And because the rectangle AE, ECis equal, as has been shewn,
to the rectangle GE, EH;

and for the same reason, the rectangle BE, ED is equal to the
same rectangle GE, EH;
therefore the rectangle AE, EC is equal to the rectangle BE, ED.
(I. ax. 1.)

Wherefore, if two straight lines, &c. q.e.d.

PROPOSITION XXXVI. THEOREM.

If from any point without a circle two straight lines be drawn, one oj
which cuts the circle, and the other touches it ; the rectangle contained by
the whole line which cuts the circle, and the part of it without the circle,
shall be equal to the square on the line which touches it.

Let B be any point without the circle ABC,

and let DC A, DB be two straight lines drawn from it,

of which DC A cuts the circle, and DB touches the same.

Then the rectangle AD, i)(7 shall be equal to tlie square on DB.

Either DC A passes through the center, or it does not:

first, let it pass through the center E,

Join EB,
therefore the angle EBD is a right angle, (ill. 18.)
And because the straight line ^ C is bisected in E, and produced
to the point D,
therefore the rectangle AD, DC, together with the square on J?C, is
equal to the square on ED : (il. 6.)

but CE is equal to EB ;
therefore the rectangle AD, DC, together with the square on EB,
is equal to the square on ED :
but the square on ED is equal to the squares on EB, BD, (i. 47.)
because EBD is a right angle :
therefore the rectangle AD, DC, together with the square on EB,
is equal to the squares on EJB, BD: (ax. 1.)

BOOK III. PROP. XXXVI. 151

take away the common square on ^B;
therefore the remaining rectangle AJD, DCis equal to the square
on the tangent DB. (ax. 3.)
Next, ii DCA does not pass through the center of the circle ABC.

Take JE the center of the circle, (ill. 1.)
draw ^J' perpendicular to ^C, (i. 12.) and join EB, EC, ED.
Because the straight line EF, which passes through the center,
cuts the straight line AC, which does not pass through the center, at
right angles ; it also bisects A C, (ill. 3.)

therefore AFh equal to EC-,

and because the straight line ^ C is bisected in F, and produced to J>,

the rectangle AD, DC, together with the square on EC,

is equal to the square on FD : (ii. 6.)

to each of these equals add the square on FE;

therefore the rectangle AD, DC, together with the squares on CF, FE,

is equal to the squares on DF, FE: (i. ax. 2.)

but the square on ED is equal to the squares on DF, FE, (l. 47.)

because EFD is a right angle ;

and for the same reason,

the square on ^Cis equal to the squares on CF, FE;

therefore the rectangle AD, DC, together with the square on EC,

is equal to the square on ED : (ax. 1.)

but CE is equal to EB ;

therefore the rectangle AD, DC, together with the square on EB,

is equal to the square on ED :

but the squares on EB, BD, are equal to the square on ED, (i. 47.)

because EBD is a right angle :

therefore the rectangle AD, DC, together with the square on EB,

is equal to the squares on EB, BD ;

take away the common square on EB ;

and the remaining rectangle AD, DC k equal to the square

on DB. (i. ax. 3.)

Wherefore, if from any point, &c. Q. E. D.
Cor. If from any point without a circle, there be drawn two straight

152 Euclid's elements.

lines cutting it, as AJB, A C, the rectangles contained by the whole
lines and the parts of them without the circle, are equal to one
another, viz. the rectangle JBA, AE, to the rectangle CA, AF: for
each of them is equal to the square on the straight line A D, which
touches the circle.

PROPOSITION XXXVII. THEOREM.

If from a point without a circle there be drawn two straight lines, one of

which cuts the circle, and the other meets it ; if the rectangle contained by the

whole line which cuts the circle, and the part of it without the circle, be equal to

the square on the line which meets it, the line which meets, shall touch the circle.

Let any point D be taken without the circle ABC,

and from it let two straight lines DCA and DB be drawn, of which

DCA cuts the circle in the points C, A, and DB meets it in

the point B.

If the rectangle AT), DC he equal to the square on DB;

then DB shall touch the circle.

D

Draw the sti'aight line DB, touching the circle ABC, in the point
D; (ill. 17.)

find F, the center of the circle, (in. 1.)

and join FF, FB, FD.

Then FFD is a right angle : (in. 18.)

and because DF touches the circle ABC, and DCA cuts it,

therefore the rectangle AD, DCis equal to the square on DF : (in. 36.)

but the rectangle AD, DC, is, by hypothesis,

equal to the square on DB :

therefore the square on DF is equal to the square on DB; (i. ax. 1.)

and the straight line DF equal to the straight line DB :

and FF is equal to FB ; (i. def. 15.)

wherefore DF, FF&ve equal to DB, BF, each to each;

and the base FD is common to the two triangles DFF, DBF;

therefore the angle DFFh equal to the angle DBF: (l. 8.)

but DFF was shewn to be a right angle ;

therefore also DBF is a right angle: (i. ax. 1.)

and BF, if produced, is a diameter ;

and the straight line which is drawn at right angles to a diameter,

from the extremity of it, touches the circle ; (in. 16. Cor.)

therefore DB touches the circle ABC.

Wherefore, if from a point, &c. Q. E. D.

NOTES TO BOOK III

In the Third Book of the Elements are demonstrated the most
"lementary properties of the circle, assuming all the properties of figures
demonstrated in the First and Second Books.

It may be worthy of remark, that the word circle will be found some-
times taken to mean the surface included within the circumference, and
sometimes the circumference itself. Euclid has employed the word (Trf.pL-
cpEpsiu) periphery, both for the whole, and for a part of the circumference
of a circle. If the word circumference were restricted to mea?i the whole
circumference, and the word arc to mean a part of it, ambiguity might
be avoided when speaking of the circumference of a circle, where only
a part of it is the subject under consideration. A circle is said to
be given in position, when the position of its center is known, and
in magnitude, when its radius is known.