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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D

E A

Of those which fall upon the concave part of the circumference

AFFC, the greatest shall be DA, which passes through the center;

128 Euclid's elements.

and any line nearer to it shall be greater than one more remote,

viz. DE shall be greater than DF, and DF greater than DC-,

but of those which fall upon the convex part of the circumference HLKG^

the least shall he DG between the point D and the diameter AG\

and any line nearer to it shall be less than one more remote,

viz. DK less than DZ, and DL less than DH.

Take i^f the center of the circle ABC, (in. 1.)

and join ME, 3IF, MC, MK, ML, MH.

And because AM is equal to 3IE,

add MD to each of these equals,

therefore ^D is equal to EM, MD : (ax. 2.)

but EM, MD are greater than ED ; (i. 20.)

therefore also AD is greater than ED.

Again, because ME is equal to MF, and MD common to the tri-

angles E3ID, F3ID; EM, MD, are equal to F3I, iÂ¥Z), each to each;

but the angle EMD is greater than the angle F3ID ; (ax. 9.)

therefore the base ED is greater than the base FD. (i. 24.)

In like manner it may be shewn that FD is greater than CD.

Therefore DA is the greatest ;

and DE greater than DF, and DF greater than DC.

And, because 3IK, KD are greater than 3ID, (i. 20.)

and MKh equal to MG, (i. def. 15.)

the remainder KD is greater than the remainder GD, (ax. 5.)

that is, GD is less than KD :

and because MLD is a triangle, and from tlie points 31, D, the

extremities of its side 3ID, the straight lines 3IK, DK are di*awn to

the point K within the triangle,

therefore 3fK, KD are less than ML, LD: (l. 21.)

but 3IK is equal to 3IL ; (i. def. 15.)

therefore, the remainder DK is less than the remainder DL. (ax. 5.)

In like manner it may be shewn, that DL is less than DH.

Therefore, DG is the least, and DK less than DL, and DL less

than Dir.

Also, there can be drawn only two equal straight lines from the

point D to the circumference, one upon each side of the line which

passes through the center.

At the point M, in the straight line MD,

make the angle D3IB equal to the angle DMK, (i. 23.) and join D5.

And because MK is equal to ME, and MD common to the tri-

angles KMD, BMD,

the two sides K3I, MD are equal to the two B3f, MD, each to each ;

and the angle K3ID is equal to the angle B3ID ; (constr.)

therefore the base DK is equal to the base DB : (l. 4 .)

but, besides DB, no straight line equal to DK can be drawn from D

to the circumference,

for, if possible, let it be DN',

and because DK is equal to DN, and also to DB,

therefore DB is equal to DN-,

that is, a L'ne nearer to the least is equal to one more remote,

which has been proved to be impossible.

K therefore, any point, &c. Q. e. d.

BOOK III. PROP. IX, X. 129

PROPOSITION IX. THEOREM.

If a point be taken within a circle, from xchich there fall more than

two equal straight lines to the circumference^ that point is the center of the

circle.

Let the point Dbe taken within the circle ^^ C, from which to the circum-

ference there fall more than two equal straight lines, viz. DA, DB, DC.

Then the point D shall be the center of the circle.

For, if not, let E, if possible, be the center :

join DE, and produce it to meet the circumference in JP, G)

then FG is a diameter of the circle ABC: (i. def. 17.)

and because in FG, the diameter of the circle ABC, there is taken

the point D, which is not the center,

therefore DG is the greatest line drawn from it to the circumference,

and DCi& greater than DB, and DB greater than DA : (iii. 7.)

but these lines are likewise equal, (hyp.) which is impossible:

therefore E is not the center of the circle ABC.

In like manner it may be demonstrated,

that no other point but D is the center ;

D therefore is the center.

Wherefore, if a point be taken, &c. Q. E. D.

PROPOSITION X. THEOREM.

One circumference of a circle cannot cut a^iother in more than two points.

If it be possible, let the circumference ABC cut the circumference

DEF in more than two points, viz. in B, G, F,

Take the center K of the circle ABC, (in. 1.) and join KB, KG, KF.

Then because jK'is the center of the circle ABC,

therefore KB, KG, KFare all equal to each other: (l. def. 15.)

and because within the circle DEF there is taken the point K, from

which to the circumference DEF fall more than two equal straight

lines KB, KG, KF;

therefore the point K is the center of the circle DEF: (ill. 9.)

but ^is also the center of the circle ABC; (constr.)

g5

130

therefore the same point is the center of two circles that cut one

another, which is impossible, (ill. 5.)

Therefore, one circumference of a circle cannot cut another in more

than two points, q.e.d.

PROPOSITION XI. THEOREM.

If one circle touch another internally in any pointy the straight line

which joins their centers being produced, shall pass through that point of

contact.

Let the circle ADEtovich the circle ^.BC internally in the point A;

and let -Pbe the center of the circle ABC, and G the center of the

circle ABU;

then the straight line which joins the centers Fy G, being produced,

shall pass through the point A.

For, i( FG produced do not pass through the point A,

let it fall otherwise, if possible, as FGJDH, and join AF, AG.

Then, because two sides of a triangle are together greater than the

third side, (i. 20.)

therefore FG, GA are greater than FA :

but FA is equal to FH; (I. def. 15.)

therefore FG, GA are greater than FS:

take away from these unequals the common part FG;

therefore the remainder AG is greater than the remainder GH; (ax. 5.)

but ^G^ is equal to GD; (i. def. 15.)

therefore GD is greater than GH,

the less than the greater, which is impossible.

Therefore the straight line which joins the points F, G, being produced,

cannot fall otherwise than upon the point A,

that is, it must ^ass through it.

Therefore, if one circle, &c. Q. E. D.

PROPOSITION XII. THEOREM.

If two circles touch each other externally in any point, the straight line

which joins their centers, shall pass through that point of contact.

Let the two circles ABC, ADE, touch each other externally in the

point A ;

and let i^be the center of the dxcleABC, and G the center o^ADE.

Then the straight line which joins the points F, G, shall pass through

the point of contact A,

BOOK III. PROP. XIII.

E

131

If not, let it pass otherwise, if possible, as FCDG, and join FA^ AG.

And because jPis the center of the circle ABC,

FA is equal to FC :

also, because G is the center of the circle ADF,

GA is equal to GD :

therefore FA, AG are equal to FC, BG-, (ax. 2.)

wherefore the whole FG is greater than FA, AG:

but FG is less than FA, AG; (I. 20.) which is impossible :

therefore the straight line which joins the points F, G, cannot pass

otherwise than through A the point of contact,

that is, FG must pass through the point A.

Therefore, if two circles, &c. Q. E. D.

PROPOSITION XIII. THEOREM.

One circle cannot touch another i7i more points than in one, whether it

touches it on the inside or outside.

For, if it be possible, let the circle FJBF touch, the circle ABC in

more points than in one.

and first on the inside, in the points B, D.

Join BD, and draw 6^-ff bisecting BD at right angles. (l. 11.)

Because the points B, D are in the circumferences of each of the circles,

therefore the straight line BD falls within each of them ; (ill. 2.)

therefore their centers are in the straight line 6^-3" which bisects BD

at right angles ; (ill. 1. Cor.)

therefore GHipasses through the point of contact: (ill. 11.)

but it does not pass through it,

because the points B, D are without the straight line GH;

which is absurd :

therefore one circle cannot touch another on the inside in more points

than in one.

Nor can two circles touch one another on the outside in more than

in one point.

For, if it be possible,

let the circle ^CX touch the circle ABC in the points A, C\

join A C.

13J2

Because tlie two points A, Care in the circumference of the circle

ACK,

therefore the straight line A C which joins them, falls within the circle

ACK: (III. 2.)

but the ciixle ACK is without the circle ABC-, (hyp.)

therefore the straight line ^ C is without this last circle :

but, because the points A, C are in the circumference of the circle ABC.

the straight line A C must be within the same circle, (ill. 2.)

which is absurd ;

therefore one circle cannot touch another on the outside in more than

in one point :

and it has been shewn, that they cannot touch on the inside in more

points than m one.

Therefore, one circle, &c. q.e.d.

PROPOSITION XIV. THEOREM.

Equal straight lines in a circle are equally distant from the center ;

and conversely, those which are equally distant from the center, are equal

to one another.

Let the straight lines AB, CD, in the circle ABDC, be equal to

one another.

Then AB and CD shall be equally distant from the center.

B

Take Ethe center of the circle ABDC, (in. 1.)

from E draw EF, EG perpendiculars to AB, CD, (i. 12.) and join

EA, EC

Then, because the straight line EF passing through the center,

cuts AB, which does not pass through the center, at right angles ;

^JP bisects AB in the point F: (ill. 3.)

therefore ^jPis equal to FB, and AB double of AF.

For the same reason CD is double of CG :

but AB is equal to CD : (hyp.)

therefore AF is equal to CG. (ax. 7.)

And because AE is equal to EC, (l. def. 15.)

the square on AE is equal to the square on EC'.

but the squares on AF, FE are equal to the square on AE, (l. 47.)

because the angle AFE is a right angle ;

BOOK III. PROP. XV. 133

and for the same reason, the squares on EG, GC are equal to the

square on EC',

therefore the squares on AF, FE are equal to the squares on CG,

KGE: (ax. 1.)

but the square on AFh equal to the square on CG,

because AF is equal to CG ;

therefore the remaining square on EF is equal to the remaining

square on EG, (ax. 3.)

and the straight line EFis therefore equal to EG:

but straight lines in a circle are said to be equally distant from the

center, when the perpendiculars di'awn to them from the center are

equal : (ill. def. 4.)

therefore AB, CD are equally distant from the center.

Conversely, let the straight lines AB, CD be equally distant from

the center, (ill. def. 4.)

that is, let FE be equal to EG ;

then AB shall be equal to CD.

For the same construction being made,

it may, as before, be demonstrated,

that AB is double of ^i^, and CD double of CG,

and that the squares on FE, AF are equal to the squares on EG, GC:

but the square on FE is equal to the square on EG,

because FE is equal to EG', (hyp.)

therefore the remaining square on AF is equal to the remaining square

on CG : (ax. 3.)

and the straight line ^i^is therefore equal to CG:

but AB was shewn to be double of AF, and CD double of CG ;

wherefore AB is equal to CD. (ax. 6.)

Therefore equal straight lines, &c. Q. e. d.

PROPOSITION XV. THEOREM.

The diameter is the greatest straight line in a circle ; and of the rest,

that which is nearer to the center is always greater than one more remote :

and conversely the greater is nearer to the center than the less.

Let ABCDhe a circle, of which the diameter is ^.D, and the center i';

and let BChe nearer to the center than EG.

Then AD shall be greater than any straight line BC, which is not a

diameter, and J5C shall be greater than EG.

A B

From E draw EH, perpendicular to BC, and ^J^to FG, (l. 12.)

and join EB, EC, EF.

And because AE is equal to EB, and ED to EC, (l. def. 15.)

therefore AD is equal to EB, EC: (ax. 2."

but EB, EC are greater than BC; (i. 20.)

wherefore also AD is greater than BC.

134

And, because BCis nearer to the center than FG, (h}-p.)

therefore JEJEis less than UK: (ill. def. 5.)

but, as was demonstrated in the preceding proposition,

^e is double of JBH, and FG double of FX,

and the squares on FH, HB are equal to the squares on FK, KF\

but the square on FH is less than the square on FK^

because FHis less than FK;

therefore the square on FH is greater than the square on FK,

and the straight line JBH greater than FK,

and therefore ^Cis greater than FG.

Next, let FChe greater than FG ;

then j5C shall be nearer to the center than FG, that is, the same con-

struction being made, FH shall be less than FK. (ill. def. 5.)

Because FCis greater than FG,

FH likewise is greater than KF:

and the squares on FH, HF are equal to the squares on FK, KF,

of which the square on FH is greater than the square on FK,

because FHis greater than FK:

therefore the square on FH is less than the square on FK,

and the straight line FH less, than FK:

and therefore FCis nearer to the center than FG. (ill. def. 5.)

â€¢ Wherefore the diameter, &c. Q. E. D.

PROPOSITION XVI. THEOREM.

The straigJit line dratcn at right angles to the diameter of a circle, from

the extremity of it, falls without the circle ; and no straight line can be drawn

from the extremitg between tJiat straight line and the circumference, so as not

to cut the circle : or, which is the same thing, no straight line can make so

great an acute angle with the diameter at its extremity, or so small an angle

with the straight line which is at right angles to it, as not to cut the circle.

Let ABChe a circle, the center of which is D, and the diameter AB.

Then the straight line drawn at right angles to AB from its ex-

tremity A, shall fall without the cii-cle.

For, if it does not, let it fall, if possible, within the circle, as ^ C;

and draw DC to the point C, where it meets the circumference.

And because DA is equal to DC, (l. def. 15.)

the angle DA C is equal to the angle A CD : (l. 5.)

but DA C is a right angle ; (hj-p.)

therefore A CD is a right angle ;

and therefore the angles DAC, A CD are equal to two right angles;

which is impossible: (l. 17.)

therefore the straight line drawn from A at right angles to BA, does

not fall within the circle.

BOOK III. PROP. XVII. 135

In the same manner it may be demonstrated,

that it does not fall upon the circumference ;

therefore it must fall without the circle, as AJS.

Also, between the straight line AJS and the circumference, no straight

line can be di-awn from the point A which does not cut the circle.

I For, if possible, let AF fall between them,

FE

and from the point D, let DGhe drawn perpendicular to AF, (l. 12.)

and let it meet the circumference in IT.

And because A GD is a right angle,

and DAG less than a right angle, (i. 17.)

therefore DA is greater than DG: (i. 19.)

but DA is equal to DH-, (i. def 15.)

therefore DJIis greater than DG,

the less than the greater, which is impossible :

â€¢ therefore no straight line can be drawn from the point A, between

AF and the circumference, which does not cut the circle :

or, which amounts to the same thing, however great an acute angle

a straight line makes with the diameter at the point A, or however

small an angle it makes with AF, the circumference must pass be-

tween that straight line and the perpendicular AF. Q.E.D.

Cor. From this it is manifest, that the straight line which is

drawn at right angles to the diameter of a circle from the extremity

of it touches the circle ; (ill. def 2.) and that it touches it only in one

point, because, if it did meet the circle in two, it would fall within it.

(ill. 2.) " Also, it is evident, that there can be but one straight line

which touches the circle in the same point."

PROPOSITION XVII. PROBLEM.

To draw a straight line from a given point, either without or in the cir-

cumference, which shall touch a given circle.

First, let ^ be a given point without the given circle FCD ;

it is required to draw a straight line from A which shall touch the circle.

â–

Find the center F of the circle, (ill. 1.) and join AF;

and from the center F, at the distance FA, describe the circle AFG ;

from the point D draw DF at right angles to FA, (i. 11.) meeting

the circumference of the circle AFG in F;

&ndjoinFBF,AJB.

136 Euclid's elements.

Then AB shall touch the circle BCD in the point B.

Beca-ise JE is the center of the circles BCD, AFG, (i. def. 15.)

therefore JEA is equal to JEF, and FD to FB ;

therefore the two sides AF, FB, are equal to the two FF, FD,

each to each :

and they contain the angle at F common to the two triangles AFB,

FFD;

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

and the triangle FFD to the triangle AFB,

and the other angles to the other angles :

therefore the angle FBA is equal to the angle FDF:

but FDF is a right angle, (constr.)

wherefore FBA is a right angle : (ax. 1.)

and FB is drawn from the center :

but a straight line drawn from the extremity of a diameter, at right

angles to it, touches the circle : (ill. 16. Cor.)

therefore AB touches the circle ;

and it is drawn from the given point A.

Secondly, if the given point be in the circumference of the circle,

as the point D,

draw DF to the center F, and DF at right angles to DF :

then Di^ touches the circle, (ill. 16. Cor.) q.e.f.

PROPOSITION XVIII. THEOREM.

If a straight line touch a circle, the straight line drawn from the center to

the point of contact, shall he perpendicular to the line touching the circle.

Let the straight line DF touch the circle ABC in the point C-,

take the center F, and di*aw the straight line FC. (iii. 1.)

Then FC shall be perpendicular to DF.

GE

It FChe not perpendicular to DF-, from the point F, if possible,

let FBG be drawn perpendicular to DF.

And because FG C is a right angle,

therefore GCFis an acute angle; (i. 17.)

and to the greater angle the greater side is opposite : (l. 19.)

therefore FC is greater than FG :

but i^Cis equal to FB ; (l. def. 15.)

therefore FB is greater than FG,

the less than the greater, which is impossible :

therefore FG is not perpendicular to DF.

In the same manner it may be shewn,

that no other line is perpendicular to DF besides FC,

that is, FC is perpendicular to DF.

Therefore, if a straight line, &c. Q. e. d.

I

BOOK I. PROP. XIX, XX. 137

PROPOSITION XIX. THEOREM.

If a straight line touch a circle, and from the point of contact a straight

line he drawn at right angles to the touching line, the center of the circle shall

be in that line.

Let the straight line DE touch the circle ABC in C,

and from C let CA be drawn at right angles to DE.

Then the center of the circle shall be in CA.

For, if not, let i^be the center, if possible, and join CF.

Because DE touches the circle ABC,

and FC is drawn from the center to the point of contact,

therefore FCh perpendicular to DE', (ill. 18.)

therefore FCE is a right angle :

but A CE is also a right angle ; (hyp.)

therefore the angle FCE is equal to the angle A CE, (ax. 1.)

the less to the greater, which is impossible :

therefore jPis not the center of the circle ABC.

In the same manner it may be shewn,

that no other point which is not in CA, is the center;

that is, the center of the circle is in CA.

Therefore, if a straight line, &c. Q. E. D.

PROPOSITION XX. THEOREM.

The angle at the center of a circle is double of the angle at the circumfer-

ence upon the same base, that is, upon the same part of the circumference.

Let ABC he a circle, and BEC an angle at the center, and BA C

an angle at the circumference, which have B C the same part of the

circumference for their base.

Then the angle BEC shall be double of the angle BA C.

A.

i

Join AE, and produce it to F.

First, let the center of the circle be within the angle BA C.

Because EA is equal to EB,

therefore the angle EBA is equal to the angle EAB ; (l. 5.)

therefore the angles EAB, EBA are double of the angle EAB :

but the angle BEFis equal to the angles EAB, EBA ; (l. 32.)

138 Euclid's elements.

therefore also the angle BJEF is double of the angle EAB-.

for the same reason, the angle FECis double of the angle EAC:

therefore the whole angle EEC is double of the whole angle BA C.

Secondly, let the center of the circle be without the angle BA C.

It may be demonstrated, as in the first case,

that the angle FECis double of the angle FAC,

and that FEB, a part of the first, is double of FAB, a part of the other ;

therefore the remaining angle BEC is double of the remaining

angle -B^ a

Therefore the angle at the center, &c. q.e.d.

PROPOSITION XXI. THEOREM.

The angles in the same segment of a circle are equal to one another.

Let A BCD be a circle,

and BAD, BED angles in the same segment BAED.

Then the angjtes BAD, BED shall be equal to one another.

First, let the segment BAED be greater than a semicircle. â™¦

A E

Take F, the center of the circle ABCD, (m. 1.) and join BE, FD

Because the angle BED is at the center, and the angle BAD al

the circumference, and that they have the same part of the circum-

ference, viz. the arc BCD for their base ;

therefore the angle BED is double of the angle BAD : (ill. 20.)

for the same reason the angle BED is double of the angle BED :

therefore the angle BAD is equal to the angle BED. (ax. 7.)

Next, let the segment BAED be not greater than a semicircle.

AE

c

Draw AF to the center, and produce it to C, and join CE.

Because ACis a diameter of the circle,

therefore the segment BADCis greater than a semicircle ;

and the angles in it BA C, BEC are equal, by the first case :

BOOK III. PROP. XXII, XXllI. 139

for the same reason, because CBED is greater than a semicircle,

the angles CAD, CED, are equal :

erefore the whole angle BAD is equal to the whole angle BED. (ax. 2.)

Wherefore the angles in the same segment, &c. Q. e. d.

PROPOSITION XXII. THEOREM.

The opposite angles of any quadrilateral figure inscribed in a circle, are

together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD.

Then any two of its opposite angles shall together be equal to two

right angles.

c

Join A C, BD.

And because the three angles of every triangle are equal to two

right angles, (I. 32.)

the three angles of the triangle CAB, viz. the angles CAB, ABCy

BCA, are equal to two right angles :

but the angle CAB is equal to the angle CDB, (ni. 21.)

because they are in the same segment CDAB;

and the angle A CB is equal to the angle ADB,

because they are in the same segment ADCB:

therefore the two angles CAB, ACB are together equal to the whole

angle ADC: (ax. 2.)

to each of these equals add the angle ABC;

therefore the three angles ABC, CAB, BCA are equal to the two

angles ABC, ADC: (ax. 2.)

but ABC, CAB, BCA, are equal to two right angles ;

therefore also the angles ABC, ADC are equal to two right angles.

In the same manner, the angles BAD, DCB, may be shewn to be

equal to two right angles.

Therefore, the opposite angles, &c. Q.E.D.

PROPOSITION XXIII. THEOREM.

Upon the same straight line, and ttpoji the same side of it, there cannot

he two similar seg7nents of circles^ not coinciding with one another.

If it be possible, upon the same straight line AB, and upon the

same side of it, let there be two similar segments of circles, A CB,

ADB, not coinciding with one another.

D

1

140 Euclid's elements.

Then, because the circumference A CB cuts the circumference AJDB

in the two points A, B, they cannot cut one another in any

other point : (ill. 10.)

therefore one of the segments must fall within the other :

let A CB fall within ADB :

draw the straight line BCD, and join CA, DA.

Because the segment A CB is similar to the segment ADB, (hyp.)

and that similar segments of circles contain equal angles ; (ill. def. 11.)

therefore the angle A CB is equal to the angle ADB,

the exterior angle to the interior, which is impossible. (l. 16.)

Therefore, there cannot be two similar segments of circles upon the

same side of the" same line, which do not coincide. Q. E. D.

PROPOSITION XXIV. THEOREM.

Similar segments of circles upon equal straight lines, are equal to one another.

Let AEB, CFD be similar segments of circles upon the equal

straight lines AB, CD.

Then the segment AEB shall be equal to the segment CFD.

E F

L \ L i

A B CD

For if the segment AEB be applied to the segment CFD,

so that the point A may be on C, and the straight line AB upon CD,

then the point B shall coincide with the point -D,

because AB is equal to CD :

therefore, the straight line AB coinciding with CD,

the segment AEB must coincide with the segment CFD, (ill. 23.)

and therefore is equal to it. (l. ax. 8.)

"Wherefore similar segments, &c. Q.E.D.

PROPOSITION XXV. PROBLEM.

A segment of a circle being given, to describe the circle of which it is the

E A

Of those which fall upon the concave part of the circumference

AFFC, the greatest shall be DA, which passes through the center;

128 Euclid's elements.

and any line nearer to it shall be greater than one more remote,

viz. DE shall be greater than DF, and DF greater than DC-,

but of those which fall upon the convex part of the circumference HLKG^

the least shall he DG between the point D and the diameter AG\

and any line nearer to it shall be less than one more remote,

viz. DK less than DZ, and DL less than DH.

Take i^f the center of the circle ABC, (in. 1.)

and join ME, 3IF, MC, MK, ML, MH.

And because AM is equal to 3IE,

add MD to each of these equals,

therefore ^D is equal to EM, MD : (ax. 2.)

but EM, MD are greater than ED ; (i. 20.)

therefore also AD is greater than ED.

Again, because ME is equal to MF, and MD common to the tri-

angles E3ID, F3ID; EM, MD, are equal to F3I, iÂ¥Z), each to each;

but the angle EMD is greater than the angle F3ID ; (ax. 9.)

therefore the base ED is greater than the base FD. (i. 24.)

In like manner it may be shewn that FD is greater than CD.

Therefore DA is the greatest ;

and DE greater than DF, and DF greater than DC.

And, because 3IK, KD are greater than 3ID, (i. 20.)

and MKh equal to MG, (i. def. 15.)

the remainder KD is greater than the remainder GD, (ax. 5.)

that is, GD is less than KD :

and because MLD is a triangle, and from tlie points 31, D, the

extremities of its side 3ID, the straight lines 3IK, DK are di*awn to

the point K within the triangle,

therefore 3fK, KD are less than ML, LD: (l. 21.)

but 3IK is equal to 3IL ; (i. def. 15.)

therefore, the remainder DK is less than the remainder DL. (ax. 5.)

In like manner it may be shewn, that DL is less than DH.

Therefore, DG is the least, and DK less than DL, and DL less

than Dir.

Also, there can be drawn only two equal straight lines from the

point D to the circumference, one upon each side of the line which

passes through the center.

At the point M, in the straight line MD,

make the angle D3IB equal to the angle DMK, (i. 23.) and join D5.

And because MK is equal to ME, and MD common to the tri-

angles KMD, BMD,

the two sides K3I, MD are equal to the two B3f, MD, each to each ;

and the angle K3ID is equal to the angle B3ID ; (constr.)

therefore the base DK is equal to the base DB : (l. 4 .)

but, besides DB, no straight line equal to DK can be drawn from D

to the circumference,

for, if possible, let it be DN',

and because DK is equal to DN, and also to DB,

therefore DB is equal to DN-,

that is, a L'ne nearer to the least is equal to one more remote,

which has been proved to be impossible.

K therefore, any point, &c. Q. e. d.

BOOK III. PROP. IX, X. 129

PROPOSITION IX. THEOREM.

If a point be taken within a circle, from xchich there fall more than

two equal straight lines to the circumference^ that point is the center of the

circle.

Let the point Dbe taken within the circle ^^ C, from which to the circum-

ference there fall more than two equal straight lines, viz. DA, DB, DC.

Then the point D shall be the center of the circle.

For, if not, let E, if possible, be the center :

join DE, and produce it to meet the circumference in JP, G)

then FG is a diameter of the circle ABC: (i. def. 17.)

and because in FG, the diameter of the circle ABC, there is taken

the point D, which is not the center,

therefore DG is the greatest line drawn from it to the circumference,

and DCi& greater than DB, and DB greater than DA : (iii. 7.)

but these lines are likewise equal, (hyp.) which is impossible:

therefore E is not the center of the circle ABC.

In like manner it may be demonstrated,

that no other point but D is the center ;

D therefore is the center.

Wherefore, if a point be taken, &c. Q. E. D.

PROPOSITION X. THEOREM.

One circumference of a circle cannot cut a^iother in more than two points.

If it be possible, let the circumference ABC cut the circumference

DEF in more than two points, viz. in B, G, F,

Take the center K of the circle ABC, (in. 1.) and join KB, KG, KF.

Then because jK'is the center of the circle ABC,

therefore KB, KG, KFare all equal to each other: (l. def. 15.)

and because within the circle DEF there is taken the point K, from

which to the circumference DEF fall more than two equal straight

lines KB, KG, KF;

therefore the point K is the center of the circle DEF: (ill. 9.)

but ^is also the center of the circle ABC; (constr.)

g5

130

therefore the same point is the center of two circles that cut one

another, which is impossible, (ill. 5.)

Therefore, one circumference of a circle cannot cut another in more

than two points, q.e.d.

PROPOSITION XI. THEOREM.

If one circle touch another internally in any pointy the straight line

which joins their centers being produced, shall pass through that point of

contact.

Let the circle ADEtovich the circle ^.BC internally in the point A;

and let -Pbe the center of the circle ABC, and G the center of the

circle ABU;

then the straight line which joins the centers Fy G, being produced,

shall pass through the point A.

For, i( FG produced do not pass through the point A,

let it fall otherwise, if possible, as FGJDH, and join AF, AG.

Then, because two sides of a triangle are together greater than the

third side, (i. 20.)

therefore FG, GA are greater than FA :

but FA is equal to FH; (I. def. 15.)

therefore FG, GA are greater than FS:

take away from these unequals the common part FG;

therefore the remainder AG is greater than the remainder GH; (ax. 5.)

but ^G^ is equal to GD; (i. def. 15.)

therefore GD is greater than GH,

the less than the greater, which is impossible.

Therefore the straight line which joins the points F, G, being produced,

cannot fall otherwise than upon the point A,

that is, it must ^ass through it.

Therefore, if one circle, &c. Q. E. D.

PROPOSITION XII. THEOREM.

If two circles touch each other externally in any point, the straight line

which joins their centers, shall pass through that point of contact.

Let the two circles ABC, ADE, touch each other externally in the

point A ;

and let i^be the center of the dxcleABC, and G the center o^ADE.

Then the straight line which joins the points F, G, shall pass through

the point of contact A,

BOOK III. PROP. XIII.

E

131

If not, let it pass otherwise, if possible, as FCDG, and join FA^ AG.

And because jPis the center of the circle ABC,

FA is equal to FC :

also, because G is the center of the circle ADF,

GA is equal to GD :

therefore FA, AG are equal to FC, BG-, (ax. 2.)

wherefore the whole FG is greater than FA, AG:

but FG is less than FA, AG; (I. 20.) which is impossible :

therefore the straight line which joins the points F, G, cannot pass

otherwise than through A the point of contact,

that is, FG must pass through the point A.

Therefore, if two circles, &c. Q. E. D.

PROPOSITION XIII. THEOREM.

One circle cannot touch another i7i more points than in one, whether it

touches it on the inside or outside.

For, if it be possible, let the circle FJBF touch, the circle ABC in

more points than in one.

and first on the inside, in the points B, D.

Join BD, and draw 6^-ff bisecting BD at right angles. (l. 11.)

Because the points B, D are in the circumferences of each of the circles,

therefore the straight line BD falls within each of them ; (ill. 2.)

therefore their centers are in the straight line 6^-3" which bisects BD

at right angles ; (ill. 1. Cor.)

therefore GHipasses through the point of contact: (ill. 11.)

but it does not pass through it,

because the points B, D are without the straight line GH;

which is absurd :

therefore one circle cannot touch another on the inside in more points

than in one.

Nor can two circles touch one another on the outside in more than

in one point.

For, if it be possible,

let the circle ^CX touch the circle ABC in the points A, C\

join A C.

13J2

Because tlie two points A, Care in the circumference of the circle

ACK,

therefore the straight line A C which joins them, falls within the circle

ACK: (III. 2.)

but the ciixle ACK is without the circle ABC-, (hyp.)

therefore the straight line ^ C is without this last circle :

but, because the points A, C are in the circumference of the circle ABC.

the straight line A C must be within the same circle, (ill. 2.)

which is absurd ;

therefore one circle cannot touch another on the outside in more than

in one point :

and it has been shewn, that they cannot touch on the inside in more

points than m one.

Therefore, one circle, &c. q.e.d.

PROPOSITION XIV. THEOREM.

Equal straight lines in a circle are equally distant from the center ;

and conversely, those which are equally distant from the center, are equal

to one another.

Let the straight lines AB, CD, in the circle ABDC, be equal to

one another.

Then AB and CD shall be equally distant from the center.

B

Take Ethe center of the circle ABDC, (in. 1.)

from E draw EF, EG perpendiculars to AB, CD, (i. 12.) and join

EA, EC

Then, because the straight line EF passing through the center,

cuts AB, which does not pass through the center, at right angles ;

^JP bisects AB in the point F: (ill. 3.)

therefore ^jPis equal to FB, and AB double of AF.

For the same reason CD is double of CG :

but AB is equal to CD : (hyp.)

therefore AF is equal to CG. (ax. 7.)

And because AE is equal to EC, (l. def. 15.)

the square on AE is equal to the square on EC'.

but the squares on AF, FE are equal to the square on AE, (l. 47.)

because the angle AFE is a right angle ;

BOOK III. PROP. XV. 133

and for the same reason, the squares on EG, GC are equal to the

square on EC',

therefore the squares on AF, FE are equal to the squares on CG,

KGE: (ax. 1.)

but the square on AFh equal to the square on CG,

because AF is equal to CG ;

therefore the remaining square on EF is equal to the remaining

square on EG, (ax. 3.)

and the straight line EFis therefore equal to EG:

but straight lines in a circle are said to be equally distant from the

center, when the perpendiculars di'awn to them from the center are

equal : (ill. def. 4.)

therefore AB, CD are equally distant from the center.

Conversely, let the straight lines AB, CD be equally distant from

the center, (ill. def. 4.)

that is, let FE be equal to EG ;

then AB shall be equal to CD.

For the same construction being made,

it may, as before, be demonstrated,

that AB is double of ^i^, and CD double of CG,

and that the squares on FE, AF are equal to the squares on EG, GC:

but the square on FE is equal to the square on EG,

because FE is equal to EG', (hyp.)

therefore the remaining square on AF is equal to the remaining square

on CG : (ax. 3.)

and the straight line ^i^is therefore equal to CG:

but AB was shewn to be double of AF, and CD double of CG ;

wherefore AB is equal to CD. (ax. 6.)

Therefore equal straight lines, &c. Q. e. d.

PROPOSITION XV. THEOREM.

The diameter is the greatest straight line in a circle ; and of the rest,

that which is nearer to the center is always greater than one more remote :

and conversely the greater is nearer to the center than the less.

Let ABCDhe a circle, of which the diameter is ^.D, and the center i';

and let BChe nearer to the center than EG.

Then AD shall be greater than any straight line BC, which is not a

diameter, and J5C shall be greater than EG.

A B

From E draw EH, perpendicular to BC, and ^J^to FG, (l. 12.)

and join EB, EC, EF.

And because AE is equal to EB, and ED to EC, (l. def. 15.)

therefore AD is equal to EB, EC: (ax. 2."

but EB, EC are greater than BC; (i. 20.)

wherefore also AD is greater than BC.

134

And, because BCis nearer to the center than FG, (h}-p.)

therefore JEJEis less than UK: (ill. def. 5.)

but, as was demonstrated in the preceding proposition,

^e is double of JBH, and FG double of FX,

and the squares on FH, HB are equal to the squares on FK, KF\

but the square on FH is less than the square on FK^

because FHis less than FK;

therefore the square on FH is greater than the square on FK,

and the straight line JBH greater than FK,

and therefore ^Cis greater than FG.

Next, let FChe greater than FG ;

then j5C shall be nearer to the center than FG, that is, the same con-

struction being made, FH shall be less than FK. (ill. def. 5.)

Because FCis greater than FG,

FH likewise is greater than KF:

and the squares on FH, HF are equal to the squares on FK, KF,

of which the square on FH is greater than the square on FK,

because FHis greater than FK:

therefore the square on FH is less than the square on FK,

and the straight line FH less, than FK:

and therefore FCis nearer to the center than FG. (ill. def. 5.)

â€¢ Wherefore the diameter, &c. Q. E. D.

PROPOSITION XVI. THEOREM.

The straigJit line dratcn at right angles to the diameter of a circle, from

the extremity of it, falls without the circle ; and no straight line can be drawn

from the extremitg between tJiat straight line and the circumference, so as not

to cut the circle : or, which is the same thing, no straight line can make so

great an acute angle with the diameter at its extremity, or so small an angle

with the straight line which is at right angles to it, as not to cut the circle.

Let ABChe a circle, the center of which is D, and the diameter AB.

Then the straight line drawn at right angles to AB from its ex-

tremity A, shall fall without the cii-cle.

For, if it does not, let it fall, if possible, within the circle, as ^ C;

and draw DC to the point C, where it meets the circumference.

And because DA is equal to DC, (l. def. 15.)

the angle DA C is equal to the angle A CD : (l. 5.)

but DA C is a right angle ; (hj-p.)

therefore A CD is a right angle ;

and therefore the angles DAC, A CD are equal to two right angles;

which is impossible: (l. 17.)

therefore the straight line drawn from A at right angles to BA, does

not fall within the circle.

BOOK III. PROP. XVII. 135

In the same manner it may be demonstrated,

that it does not fall upon the circumference ;

therefore it must fall without the circle, as AJS.

Also, between the straight line AJS and the circumference, no straight

line can be di-awn from the point A which does not cut the circle.

I For, if possible, let AF fall between them,

FE

and from the point D, let DGhe drawn perpendicular to AF, (l. 12.)

and let it meet the circumference in IT.

And because A GD is a right angle,

and DAG less than a right angle, (i. 17.)

therefore DA is greater than DG: (i. 19.)

but DA is equal to DH-, (i. def 15.)

therefore DJIis greater than DG,

the less than the greater, which is impossible :

â€¢ therefore no straight line can be drawn from the point A, between

AF and the circumference, which does not cut the circle :

or, which amounts to the same thing, however great an acute angle

a straight line makes with the diameter at the point A, or however

small an angle it makes with AF, the circumference must pass be-

tween that straight line and the perpendicular AF. Q.E.D.

Cor. From this it is manifest, that the straight line which is

drawn at right angles to the diameter of a circle from the extremity

of it touches the circle ; (ill. def 2.) and that it touches it only in one

point, because, if it did meet the circle in two, it would fall within it.

(ill. 2.) " Also, it is evident, that there can be but one straight line

which touches the circle in the same point."

PROPOSITION XVII. PROBLEM.

To draw a straight line from a given point, either without or in the cir-

cumference, which shall touch a given circle.

First, let ^ be a given point without the given circle FCD ;

it is required to draw a straight line from A which shall touch the circle.

â–

Find the center F of the circle, (ill. 1.) and join AF;

and from the center F, at the distance FA, describe the circle AFG ;

from the point D draw DF at right angles to FA, (i. 11.) meeting

the circumference of the circle AFG in F;

&ndjoinFBF,AJB.

136 Euclid's elements.

Then AB shall touch the circle BCD in the point B.

Beca-ise JE is the center of the circles BCD, AFG, (i. def. 15.)

therefore JEA is equal to JEF, and FD to FB ;

therefore the two sides AF, FB, are equal to the two FF, FD,

each to each :

and they contain the angle at F common to the two triangles AFB,

FFD;

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

and the triangle FFD to the triangle AFB,

and the other angles to the other angles :

therefore the angle FBA is equal to the angle FDF:

but FDF is a right angle, (constr.)

wherefore FBA is a right angle : (ax. 1.)

and FB is drawn from the center :

but a straight line drawn from the extremity of a diameter, at right

angles to it, touches the circle : (ill. 16. Cor.)

therefore AB touches the circle ;

and it is drawn from the given point A.

Secondly, if the given point be in the circumference of the circle,

as the point D,

draw DF to the center F, and DF at right angles to DF :

then Di^ touches the circle, (ill. 16. Cor.) q.e.f.

PROPOSITION XVIII. THEOREM.

If a straight line touch a circle, the straight line drawn from the center to

the point of contact, shall he perpendicular to the line touching the circle.

Let the straight line DF touch the circle ABC in the point C-,

take the center F, and di*aw the straight line FC. (iii. 1.)

Then FC shall be perpendicular to DF.

GE

It FChe not perpendicular to DF-, from the point F, if possible,

let FBG be drawn perpendicular to DF.

And because FG C is a right angle,

therefore GCFis an acute angle; (i. 17.)

and to the greater angle the greater side is opposite : (l. 19.)

therefore FC is greater than FG :

but i^Cis equal to FB ; (l. def. 15.)

therefore FB is greater than FG,

the less than the greater, which is impossible :

therefore FG is not perpendicular to DF.

In the same manner it may be shewn,

that no other line is perpendicular to DF besides FC,

that is, FC is perpendicular to DF.

Therefore, if a straight line, &c. Q. e. d.

I

BOOK I. PROP. XIX, XX. 137

PROPOSITION XIX. THEOREM.

If a straight line touch a circle, and from the point of contact a straight

line he drawn at right angles to the touching line, the center of the circle shall

be in that line.

Let the straight line DE touch the circle ABC in C,

and from C let CA be drawn at right angles to DE.

Then the center of the circle shall be in CA.

For, if not, let i^be the center, if possible, and join CF.

Because DE touches the circle ABC,

and FC is drawn from the center to the point of contact,

therefore FCh perpendicular to DE', (ill. 18.)

therefore FCE is a right angle :

but A CE is also a right angle ; (hyp.)

therefore the angle FCE is equal to the angle A CE, (ax. 1.)

the less to the greater, which is impossible :

therefore jPis not the center of the circle ABC.

In the same manner it may be shewn,

that no other point which is not in CA, is the center;

that is, the center of the circle is in CA.

Therefore, if a straight line, &c. Q. E. D.

PROPOSITION XX. THEOREM.

The angle at the center of a circle is double of the angle at the circumfer-

ence upon the same base, that is, upon the same part of the circumference.

Let ABC he a circle, and BEC an angle at the center, and BA C

an angle at the circumference, which have B C the same part of the

circumference for their base.

Then the angle BEC shall be double of the angle BA C.

A.

i

Join AE, and produce it to F.

First, let the center of the circle be within the angle BA C.

Because EA is equal to EB,

therefore the angle EBA is equal to the angle EAB ; (l. 5.)

therefore the angles EAB, EBA are double of the angle EAB :

but the angle BEFis equal to the angles EAB, EBA ; (l. 32.)

138 Euclid's elements.

therefore also the angle BJEF is double of the angle EAB-.

for the same reason, the angle FECis double of the angle EAC:

therefore the whole angle EEC is double of the whole angle BA C.

Secondly, let the center of the circle be without the angle BA C.

It may be demonstrated, as in the first case,

that the angle FECis double of the angle FAC,

and that FEB, a part of the first, is double of FAB, a part of the other ;

therefore the remaining angle BEC is double of the remaining

angle -B^ a

Therefore the angle at the center, &c. q.e.d.

PROPOSITION XXI. THEOREM.

The angles in the same segment of a circle are equal to one another.

Let A BCD be a circle,

and BAD, BED angles in the same segment BAED.

Then the angjtes BAD, BED shall be equal to one another.

First, let the segment BAED be greater than a semicircle. â™¦

A E

Take F, the center of the circle ABCD, (m. 1.) and join BE, FD

Because the angle BED is at the center, and the angle BAD al

the circumference, and that they have the same part of the circum-

ference, viz. the arc BCD for their base ;

therefore the angle BED is double of the angle BAD : (ill. 20.)

for the same reason the angle BED is double of the angle BED :

therefore the angle BAD is equal to the angle BED. (ax. 7.)

Next, let the segment BAED be not greater than a semicircle.

AE

c

Draw AF to the center, and produce it to C, and join CE.

Because ACis a diameter of the circle,

therefore the segment BADCis greater than a semicircle ;

and the angles in it BA C, BEC are equal, by the first case :

BOOK III. PROP. XXII, XXllI. 139

for the same reason, because CBED is greater than a semicircle,

the angles CAD, CED, are equal :

erefore the whole angle BAD is equal to the whole angle BED. (ax. 2.)

Wherefore the angles in the same segment, &c. Q. e. d.

PROPOSITION XXII. THEOREM.

The opposite angles of any quadrilateral figure inscribed in a circle, are

together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD.

Then any two of its opposite angles shall together be equal to two

right angles.

c

Join A C, BD.

And because the three angles of every triangle are equal to two

right angles, (I. 32.)

the three angles of the triangle CAB, viz. the angles CAB, ABCy

BCA, are equal to two right angles :

but the angle CAB is equal to the angle CDB, (ni. 21.)

because they are in the same segment CDAB;

and the angle A CB is equal to the angle ADB,

because they are in the same segment ADCB:

therefore the two angles CAB, ACB are together equal to the whole

angle ADC: (ax. 2.)

to each of these equals add the angle ABC;

therefore the three angles ABC, CAB, BCA are equal to the two

angles ABC, ADC: (ax. 2.)

but ABC, CAB, BCA, are equal to two right angles ;

therefore also the angles ABC, ADC are equal to two right angles.

In the same manner, the angles BAD, DCB, may be shewn to be

equal to two right angles.

Therefore, the opposite angles, &c. Q.E.D.

PROPOSITION XXIII. THEOREM.

Upon the same straight line, and ttpoji the same side of it, there cannot

he two similar seg7nents of circles^ not coinciding with one another.

If it be possible, upon the same straight line AB, and upon the

same side of it, let there be two similar segments of circles, A CB,

ADB, not coinciding with one another.

D

1

140 Euclid's elements.

Then, because the circumference A CB cuts the circumference AJDB

in the two points A, B, they cannot cut one another in any

other point : (ill. 10.)

therefore one of the segments must fall within the other :

let A CB fall within ADB :

draw the straight line BCD, and join CA, DA.

Because the segment A CB is similar to the segment ADB, (hyp.)

and that similar segments of circles contain equal angles ; (ill. def. 11.)

therefore the angle A CB is equal to the angle ADB,

the exterior angle to the interior, which is impossible. (l. 16.)

Therefore, there cannot be two similar segments of circles upon the

same side of the" same line, which do not coincide. Q. E. D.

PROPOSITION XXIV. THEOREM.

Similar segments of circles upon equal straight lines, are equal to one another.

Let AEB, CFD be similar segments of circles upon the equal

straight lines AB, CD.

Then the segment AEB shall be equal to the segment CFD.

E F

L \ L i

A B CD

For if the segment AEB be applied to the segment CFD,

so that the point A may be on C, and the straight line AB upon CD,

then the point B shall coincide with the point -D,

because AB is equal to CD :

therefore, the straight line AB coinciding with CD,

the segment AEB must coincide with the segment CFD, (ill. 23.)

and therefore is equal to it. (l. ax. 8.)

"Wherefore similar segments, &c. Q.E.D.

PROPOSITION XXV. PROBLEM.

A segment of a circle being given, to describe the circle of which it is the

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