Euclid.

# A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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Online LibraryEuclidA text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 11 of 27)
Font size .'. the ^ FED is a rt. angle.
.-. the^ FEC=ithe /. FED,
the whole equal to its part, which is impossible.
.'.AC and BD are not hoth bisected at E. Q. e. d.

[For Exercises, see page 158.] /.

iab KKCi.iDs ki;kmknt!

/ Proposition 5. Theorem.

If two circles cut one another^ they cannot have the same
centre.

Let the two 0^ AGC, BFC cut one another at C;
then they shall not have the same centre.
For, if possible, let the two circles have the same centre;
and let it be called E.
Join EC;
and from E draw any st. line to meet the O''^^ at F and G.
Then, because E is the centre of the 0AGC, Hyp.
.'. EG = EC.
And because E is also the centre of the BFC, JI^p.
.-. EF=EC.
.-. EG = EF,
the whole equal to its part, which is impossible.
.*. the two circles have not the same centre.

Q. E. D.

EXERCISES.
ON Proposition 4.

1. If a parallelogram can be inscribed in a circle, the point of
intersection of its diagonals must be at the centre of the circle.

2. Rectangles are the only parallelograms that can be inscribed
in a circle.

ON Proposition 5.

3. Two circles, which intersect at one point, must also intersect
at another.

BOOK in. riiOF. 6. 159

Proposition 6. Theorem.

If two circles touch one another internally^ they cannot
have the same ceritre.

Let the two Â©" ABC, DEC touch one another internally
at C:

then they shall not have the same centre.

For, if possible, let the two circles have the same centre ;
and let it be called F.

Join FC;
and from F draw any st. line to meet the O ^*^^ at E and B.

Then, because F is the centre of the 0ABC, Hyp.
.'. FB=FC.

I

And because F is the centre of the DEC, Hy])
.'. FE=FC.

.-. FB=FE;

the whole equal to its part, which is impossible.

the two circles have not the same centre, q. E.i>.

Note. From Propositions 5 and 6 it is seen that circles, whose
circumferences have any point in common, cannot be concentric,
unless they coincide entirely.

Conversely, the circumferences of concentric circles can have no
point in common.

160 KUCl.lh S Kl.K.MENTS.

Proposition 7. Theorem.

If from any 'point within a circle which is not the centre,
straight lines are drawn to the circumference, the greatest is
that which passes through the centre ; and the least is that
which, when produced backwards, passes thorough the centre :

and of all other such lines, that which is nearer to the
greatest is always greater than one more remote :

also two equal straight lines, and only two, can, be drawn
from the given point to the circumference, one on eacli side
of the diameter.

Let A BCD be a circle, within which any point F is taken,
which is not the centre: let FA, FB, FC, FG be drawn to
the 0^^% of which FA passes through E the centre, and FB is
nearer than FC to FA, and FC nearer than FG : and let
FD be the line which, when produced backwards, passes
through the centre : then of all these st. lines
(i) FA shall be the greatest;
(ii) FD shall be the least;

(iii) FB shall be greater than FC, and FC greater
than FG;

(iv) also two, and only two, equal st. lines can Ix;
drawn from F to tlie O"^*^.

Join EB, EC, EG.

(i) Then in the A FEB, the two sides FE, EB are together

greater than the third side FB. i. 20.

But EB = EA, being radii of the circle;

.'. FE, EA are together greater than FB;

that is, FA is greater than FB.

BOOK III. PROP. 7. 161

Similarly FA may be shewn to be greater than any other
st. line drawn from F to the O'^'^;

.'. FA is the greatest of all such lines.

(ii) In the A EFG, the two sides EF, FG are together
greater than EG ; I. 20.

and EG = ED, being radii of the circle;
.*. EF, FG are together greater than ED.
Take away the common part EF;
then FG is greater than FD.
Similarly any other st. line drawn from F to the O^Â®
may be shewn to be greater than FD.

.'. FD is the least of all such lines.

(iii) In the A^ BEF, CEF,

( BE=:CE, III. Z)^/ 1.

Because â– < and EF is common;

(but the z. BEF is greater than the z. CEF;

.'. FB is greater than FG. i. 24.

Similarly it may be shewn that FC is greater than FG.

(iv) At E in FE make the lFEH equal to the z. FEG.

I. 23.
Join FH.
Then in the AÂ« GEF, HEF,

r GE= HE, III. Def. 1.

Because â€˘< and EF is common;

(also the l GEF = the z. HEF; Constr.
.'. FG = FH. I. 4.

And besides FH no other straight line can be drawn
rom F to the O^Â® equal to FG.

For, if possible, let FK= FG.

Then, because FH = FG, Proved.

.-. FK=: FH,
that is, a line nearer to FA, the greatest, is equal to a line
Avhich is more remote; which is impossible. Proved.

.'. two, and only two, equal st. lines can be drawn from
FtotheO'''. Q.E.D.

H. E. 11

162 kucltd s elemknts.

Proposition 8. Theorem.

If from any point witliout a circle straight lines are drawn
to the circumference^ of those which fall on the concave cir-
cumference, tlie greatest is that which passes through the
centre ;) and of others, that which is nearer to the greatest
is always greater than one more remote :

further, of those which fall on the convex circumference,
the least is that which, when produced, passes through the
centre ;\ and of others that which is nearer to the least is
always less than one more remote:

lastly, from the given point there can he drawn to the
circumference two, and only two, equal straight lines, one on
each side of the shoi'test line.

Let BG D be a circle of which C is the centre ; and let
A be any point outside the circle : let ABD, AEH, AFG, be
St. lines drawn from A, of which AD passes through C, the
centre, and AH is nearer than AG to AD :

then of St. lines drawn from A to the concave O ^^,
(i) AD shall be the greatest, and (ii) AH greater than
AG :

and of st. lines drawn from A to the convex O '^^
(iii) AB shall be the least, and (iv) AE less than AF.
(v) Also two, and only two, equal st. lines can be
drawn from A to the O*^^.

Join CH, CG, CF, CE.
(i) Then in the A ACH, the two sides AC, CH are
together greater than AH : i. 20.

but CH = CD, being radii of the circle;
.'. AC, CD are together greater than AH:
that is, AD is greater than AH.
Similarly AD may be shewn to be greater than any other
st. line drawn from A to the concave O^'';

.'. AD is the greatest of all such lines.

BOOK III. PROP. 8. 163

(ii) In the AÂ« HCA, GCA,

( HC = GC, 111. Be/. 1.

Because â– < and CA is common;

( but the L HCA is greater than the z. GCA;
.*. AH is greater than AG. i. 24.

(iii) In the A A EC, the two sides AE, EC are together

greater than AC : i. 20.

but EC=BC; iii. Def. 1.

.'. the remainder AE is greater than the remainder AB.

Similarly any other st. line drawn from A to the convex

O^Â® may be shewn to be greater than AB;

.'. AB is the least of all such lines.

(iv) In the A AFC, because AE, EC are drawn from the
extremities of the base to a point E within the triangle,

,". AF, FC are together greater than AE, EC. i. 21.
But FC = EC, III. Def, 1.

.*. the remainder AF is greater than the remainder AE.

(v) At C, in AC, make the l ACM equal to the l ACE.
Join AM.
Then in the two AÂ« EC A, MCA,

EC=MC, III. Def. 1.

Because â– { and CA is common ;

also the l ECA = the /.MCA; Constr.
.â€˘.AE = AM; I. 4.

and besides AM, no st. line can be drawn from A to the
0^^% equal to AE.

For, if possible, let AK = AE :
then because AM = AE, Proved.

AM = AK;

that is, a line nearer to the shortest line is equal to a
line which is more remote ; which is impossible. Proved.
.'. two, and only two, equal st. lines can be drawn from
A to the 0'=*'. Q.E.D.

Where are the limits of that part of the circumference which is
concave to the point A ?

11-2

i(j4 EUCLID'S KLEMKNTS.

Ohs. Of tlie following proposition Euclid gave two distinct pioofV
the first of which has the advantage of being direct.

Proposition 9. Theorem. [First Proof.]

Jf from, a 'point within a circle more tJtan two equal
straight lines can he drawn to the circumference, that point
is the centre of the circle.

Let ABC be a circle, and D a point within it, from which
more than two equal st. lines are drawn to the O "''5 namely
DA, DB, DC :

then D shall be the centre of the circle.

Join AB, BC :
and bisect AB, BC at E and F respectively. i. 10.
Join DE, DF.
Then in the AÂ« DEA, DEB,

f EA = EB, Consir.

Because < and DE is common;

[ and DA=^ DB; Hyp.

:. the L DEA = the l DEB; I. 8.

.'. these angles, being adjacent, are rt. angles.

Hence ED, which bisects the chord AB at rt. angles, must

pass through the centre. iii. 1. Cor.

Similarly it may be shewn that FD passes through the

centre.

.'. D, which is the only point common to ED and FD,
must be the centre. q.e.d.

BOOK III. PROF. 9. ' 165

Proposition 9. Theorem. [Second Proof.]

If froin a point loithin a circle more than two equal
straiyht lijies can be drawn to the circumference, that imint
is the centre of the circle.

Let ABC be a circle, and D a point within it, from which
more than two equal st. lines are drawn to the O^^ namely
DA, DB, DC:

then D shall be the centre of the circle.

For, if not, suppose E to be the centre.
Join DE, and produce it to meet the O^Â® at F, G.

Then because D is a point within the circle, not the
centre, and because DF passes through the centre E ;

.'. DA, which is nearer to DF, is greater than DB, whicli
is more remote : iii. 7.

but this is impossible, since by liypothesis, DA, DB, are
equal.

.â– . E is not the centre of the circle.

*And wherever we suppose the centre E to be, other-
wise than at D, two at least of the st. lines DA, DB, DC
may be shewn to be unequal, which is contrary to hypo-
thesis.

.'. D is the centre of the ABC.

Q.E.D.

* Note. For example, if the centre E were supposed to be within
the angle BDC, then DB would be greater than DA; if within the
angle ADB, then DB would be greater than DC ; if on one of the three
straight lines, as DB, then DB would be greater than both DA and DC.

166 EUCLID'S ELEMENTS.

Ohs. Two proofs of Proposition 10, both indirect, were given by
Euclid.

Proposition 10. Theorem. [First Proof.]
One circle cannot cut anotlier at more than two poirUs.

If possible, let DABC, EABC be two circles, cutting one
another at more than two points, namely at A, B, C.
Join AB, BC.
Draw FH, bisecting AB at rt. angles; i. 10, 11.

and draw GH bisecting BC at rt. angles.
Then because AB is a chord of hath circles, and FH
bisects it at vi. angles,

.'. the centre of both circles lies in FH. iii. l.Cor.
Again, because BC is a chord of both circles, and GH
bisects it at right angles,

.'. the centre of both circles lies in GH. iii. l.Cor.
Hence H, the only point common to FH and GH, is the
centre of both circles ;

which is impossible, for circles which cut one another

cannot have a common centre. iii. 5.

.*. one circle cannot cut another at more than two points.

Q.E.D.

Corollaries, (i) Two circles cannot meet in three
points vnthout coinciding entirely.

(ii) Two circles cannot have a common arc ivithout
coinciding entirely.

(iii) Ordy one circle can he described through three
poifits, which are not in the same straight line.

BOOK III. PROP. 10. 167

Proposition 10. Theorem. [Second Proof.]

Oiie circle cmmot cut another at more than two joints.
D

If possible, let DABC, EABC be two circles, cutting one
another at more than two points, namely at A, B, C.

Find H, the centre of the DABC, in. 1.

and join HA, HB, HC.
Then since H is the centre of the Â© DABC,

.'. HA, HB, HC are all equal. in. Def. 1.

And because H is a point within the Â© EABC, from
which more than two equal st. lines, namely HA, HB, HC
are drawn to the O*^Â®,

.*. H is the centre of the Â© EABC : iii. 9.

that is to say, the two circles have a common centre H ;

but this is impossible, since they cut one another, in. 5.

Therefore one circle cannot cut another in more than

two points. Q.E.D.

Note. This proof is imperfect, because it assumes that the centre
of the circle DABC must fall within the circle EABC; whereas it
may he conceived to fall either without the circle EABC, or on its
circumference. Hence to make the proof complete, two additional
cases are required.

168 euclid's elements.

Proposition 11. Theorem.

If two circles touch one another internally, the straiglU
line which joins their centres^ being produced, shall ]>ass
through the point of contact.

Let ABC and ADE be two circles which touch one
another internally at A ; let F be the centre of the ABC,
and G the centre of the ADE:

then shall FG produced pass through A.

If not, let it pass otherwise, as FGEH.
Join FA, GA.
Then in the A FGA, the two sides FG, Q'A are together
greater than FA : i. 20.

but FA ^ FH, being radii of the ABC : â€˘ Hyp.
.'. FG, GA are together greater than FH.
Take away the common part FG ;
then GA is greater than GH.
.'. GE is greater than GH,
the part greater than the whole ; which is impossible.
.*. FG, when produced, must pass through A.

Q.E.D.
EXERCISES.

1. If the distance between the centres of two circles is equal to
the difference of their radii, then the circles must meet in one point,
but in no other; that is, they must touch one another.

2. If two circles whose centres are A and B touch onr another
internally, and a straight line be drawn through their point of contact,
cutting the circumferences at P and Q; shew that the radii AP and BQ
are parallel.

book iii. prop. 12. 169

Proposition 12. Theorem.

If two circles touch one another externallij, the straight
line which joins their centres shall pass through the jjoint of
contact.

â–

Let ABC and ADE be two circles which touch one
another externally at A; let F be the centre of the ABC,
and G the centre of the ADE :

then shall FG pass through A.

If not, let FG pass otherwise, as FHKG.

Join FA, GA.

Then in the A FAG, the two sides FA, GA are together
greater than FG : i. 20.

but FA ^ FH, being radii of the ABC ; Hyp-

.". FH and GK are together greater than FGj
which is impossible.
.'. FG must pass through A.

Q.E.D.

EXERCISES.

1. ÂĄind the locus of the centres of all circles ichich touch a (jiven
circle at a given point.

2. Find the locus of the centres of all circles of given radius, xohich
touch a given circle.

3. If the distance between the centres of two circles is equal to
the sum of their radii, then the circles meet in one point, but in no
other; that is, they touch one another.

4. If two circles whose centres are A and B touch one another
externally, and a straight line be drawn through their point of contact
cutting the circumferences at P and Ql; shew that the radii AP and BQ
arc parallel.

170

Euclid's elements.

Proposition 13. Theorem.

Ttuo circles cannot touch one another at more than one
pointj whether internally or externally.

Fig. 2

D G

If possible, let ABC, EDF be two circles which touch one
another at more than one point, namely at B and D.

Join BD;
and draw GF, bisecting BD at rt. angles, i. 10, 11.

Then, whether the circles touch one another internally,
as in Fig. 1, or externally as in Fig. 2,

because B and D are on the O ^^^ of both circles,
.'. BD is a chord of both circles :
.'. the centres of both circles lie in GF, which bisects BD

III. 1. Cor.

Hence GF which joins the centres must pass through

a point of contact; iii. 11, and 12.

which is impossible, since B and D are without GF.

.*. two circles cannot touch one another at more than

one point.

Q.E.D.

at rt. angles.

Note. It must be observed that the proof here given applies, by
virtue of Propositions 11 and 12, to both the above figures: we have
therefore omitted the separate discussion of Fig. 2, which finds a place
in most editions based on Simson's text.

BOOK III. PROP. 13. 171

EXERCISES ON PROPOSITIONS 1â€”13.

1. Describe a circle to pass through two given points and have
its centre on a given straight line. When is this impossible ?

2. All circles which pass through a fixed point, and have their
centres on a given straight line, pass also through a second fixed
point.

3. Describe a circle of given radius to touch a given circle at a
given point. How many solutions will there be? When will there
be only one solution?

4. From a given point as centre describe a circle to touch a given
circle. How many solutions will there be?

5. Describe a circle to pass through a given point, and touch a
given circle at a given point. [See Ex. 1, p. 169 and Ex. 5, p. 156.]
When is this impossible?

6. Describe a circle of given radius to touch two given circles,
[See Ex. 2, p. 169.] How many solutions will there be ?

7. Two parallel chords of a circle are six inches and eight inches
in length respectively, and the perpendicular distance between them
is one inch : find the radius.

8. If two circles touch one another externally, the straight lines,
which join the extremities of parallel diameters towards opposite
parts, must pass through the point of contact.

9. Find the greatest and least straight lines which have one
extremity on each of two given circles, which do not intersect.

10. In any segment of a circle, of all straight lines drawn at right
angles to the chord and intercepted between the chord and the arc,
the greatest is that which passes through the middle point of the
chord ; and of others that which is nearer the greatest is greater than
one more remote.

11. If from any point on the circumference of a circle straight
lines be drawn to the circumference, the greatest is that which passes
through the centre ; and of others, that which is nearer to the greatest
is greater than one more remote ; and from this point there can be
drawn to the circumference two, and only two, equal straight lines.

172 euclid's elkments.

Proposition 14. Theorem.

Equal chords in a circle are equidistant from the centre:
and, conversely, chords which are equidistant from the
centre are equal.

A

Let ABC be a circle, and let AB and CD be cliords,
of wliicli the perp. distances from the centre are EF
and EG.

First, Let AB = CD :

then shall AB and CD be equidistant from the centre E.
Join EA, EC.
Then, because EF, which passes through the centre, is
perp. to the chord AB; Hyp-

:. EF bisects AB ; ill. 3.

that is, AB is double of FA.
For a similar reason, CD is double of GC.

ButAB = CD, Uyi>.

.'. FA = GC.
Now EA = EC, being radii of tlie circle;

.'. the sq. on EA = the sq. on EC.
But the sq. on EA = the sqq. on EF, FA;

for the L EFA is a rt. angle. i. 47.

And the sq. on EC = the sqq. on EG, GC ;

for the z_ EGC is a rt. angle.

.'. the sqq. on EF, FA = the sqq. on EG, GC.

Now of these, the sq. on FA = the sq. on GC ; for FA = GC.

.*. the sq. on EF = the sq. on EG,

.-. EF=EG;

that is, the chords AB, CD are equidistant from the centre.

Q.E.D.

I

BOOK III. PROP. 14. 173

Conversely. Let AB and CD be equidistant from the
centre E ;

that is, let EF - EG :
then shall AB â€” CD.
For, the same construction being made, it may be
shewn as before that AB is double of FA, and CD double
of GC;

and that the sqq. on EF, FA = the sqq. on EG, GC.
Now of these, the sq. on EF = the sq. on EG,

for EF = EG : Hyp.

.'. the sq. on FA = the square on GC ;

.-. FA = GC;

and doubles of these equals are equal ;

that is, AB = CD.

Q.E.D.

EXERCISES.

1. Find the locus of the middle points of equal chords of a circle.

2. If two chords of a circle cut one another, and make equal
angles with the straight line which joins their point of intersection
to the centre, they are equal.

3. If two equal chords of a circle intersect, shew that the segments
of the one are equxil respectively to the segments of the other.

4. In a given circle draw a chord which shall be equal to one
given straight line (not greater than the diameter) and parallel to
another.

5. PQ is a fixed chord in a circle, and AB is any diameter: shew
that the difference of the perpendiculars let fall from A and B on PQ
is constant, that is, the same for all positions of AB.

174 euclid's elements.

Proposition 15. Theorem.

The diameter is the greatest chord in a circle ;

and of otherSy that which is nearer to the centre is greater
than one more remote:

conversely^ the greater chord is nearer to the centre titan
the less.

Let A BCD be a circle, of which AD is a diameter, and E
the centre ; and let BC and FG be any two chords, whoso
perp. distances from the centre are EH and EK :
then (i) AD shall be greater than BC :

(ii) if EH is less than EK, BC shall be greater than FG :
(iii) if BC is greater than FG, EH shall be less than EK.

(i) Join EB, EC.

Then in the A BEC, the two sides BE, EC are together
greater than BC ; i. 20.

but BE = AE, III. Def. 1.

and EC = ED ;
â€˘ .*. AE and ED together are greater than BC ;

that is, AD is greater than BC.
Similarly AD may be shewn to be greater than any
other chord, not a diameter.

(ii) Let EH be less than EK;

then BC shall be greater than FG.
Join EF.
Since EH, passing through the centre, is perp. to the
chord BC,

.'. EH bisects BC ; iii. 3.

that is, BC is double of HB.
For a similar reason FG is double of KF.

Now EB= EF, III. Bef. 1.

.*. the sq. on EB = the sq. on EF.
But the sq. on EB = the sqq. on EH, HB;

for the z. EHB is a rt. angle ; i. 47.

also the sq. on EF = the sqq. on EK, KF;
for the z_ EKF is a rt. angle.
.'. the sqq. on EH, HB = the sqq. on EK, KF.
But the sq. on EH is less than the sq. on EK,

for EH is less than EK ; J^yP'

.'. the sq. on HB is greater than the sq. on KF ;
.*. HB is greater than KF :
hence BC is greater than FG.

(iii) Let BC be greater than FG ;

then EH shall be less than EK.
For since BC is greater than FG, JiyV-

.'. HB is greater than KF :
.". the sq. on HB is greater than the sq. on KF.
But the sqq. on EH, HB=:the sqq. on EK, KF : Proved.
.'. the sq. on EH is less than the sq. on EK ;
.'. EH is less than EK.

Q.E.D.

EXERCISES.

1. Through a given point within a circle draw the least possible
chord.

2. AB is a fixed chord of a circle, and XY any other chord having
its middle point Z on AB: what is the greatest, and what the least
length that XY may have?

Shew that XY increases, as Z approaches the middle point of AB.

3. In a given circle draw a chord of given length, having its
middle point on a given chord.

When is this problem impossible?

176

EUCLID'S ELEMENTS.

01)8. Of the following proofs of Proposition IG, the second (by
reductio ad absurdura) is that given by Euclid ; but the first is to be
preferred, as it is direct^ and not less simple than the other.

Proposition 16. Theorem. [Alternative Proof.]

The straight line drawn at right angles to a diameter of
a circle at one of its extremities is a tangent to the circle:

and every other straight line drawn through this point
cuts the circle.

Let AKB be a circle, of which E is the centre, and AB
a diameter; and through B let thie st. line CBD be drawn
at rt. angles to AB :

then (i) CBD shall be a tangent tO the circle;

(ii) any other st. line through B, as BF, shall cut
the circle.

(i) In CD take any point G, and join EG.

Then, in the AEBG, the l EBG is a rt. angle; Hyp.
.'. the z. EGB is less than a rt. angle; i. 17.

.". the L EBG is greater than the z. EGB;

.'. EG is greater than EB: l. 19.

that is, EG is greater than a radius of tlie circle;
.'. tlie point G is without the circle.
Similarly any other point in CD, except B, may be shewn
to be outside the circle :

hence CD meets the circle at B, but being produced,
does not cut it;

that is, CD is a tangent to the circle. iii.Z>^10.

BOOK III. PROP. 16. 177

(ii) Draw EH perp. to BF. i. 12.
Then in the A EH B, because the ^ EHB is a rt. angle,

.'. the lEBH is less than a rt. angle; i. 17.

.". EB is greater than EH; I. 19.
that is, EH is less than a radius of the circle :
.". H, a point in BF, is within the circle;

.*. BF must cut the circle. q.e.d.

Proposition 16. Theorem. [Euclid's Proof.]

The straight line drawn at right angles to a diameter of
a circle at one of its extremities, is a tangent to the circle :

and no other straight line can be drawn through this
point so as not to cut the circle.

E

Let ABC be a circle, of which D is the centre, and AB
a diameter; let AE be drawn at rt. angles to BA, at its
extremity A:

(i) then shall AE be a tangent to the circle.
For, if not, let AE cut the circle at C.

Join DC.
Then in the A DAC, because DA = DC, ill. Bef 1.

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