BOOK III.
DEFINITIONS.
1
Equal circles are those the diameters of which are equal, or the radii of which are equal.
2
A straight line is said to
touch a circle which, meeting the circle and being produced, does not cut the circle.
3
Circles are said to
touch one another which, meeting one another, do not cut one another.
4
In a circle straight lines are said
to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal.
5
And that straight line is said to be
at a greater distance on which the greater perpendicular falls.
6
A
segment of a circle is the figure contained by a straight line and a circumference of a circle.
7
An
angle of a segment is that contained by a straight line and a circumference of a circle.
8
An
angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the
base of the segment, is contained by the straight lines so joined.
9
And, when the straight lines containing the angle cut off a circumference, the angle is said to
stand upon that circumference.
10
A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.
11
Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another.
BOOK III. PROPOSITIONS.
PROPOSITION 1.
To find the centre of a given circle.
Let
ABC be the given circle; thus it is required to find the centre of the circle
ABC.
Let a straight line
AB be drawn
through it at random, and let it be bisected at the point
D; from
D let
DC be drawn at right angles to
AB and let it be drawn through to
E; let
CE be bisected at
F;
I say that
F is the centre of the circle
ABC.
For suppose it is not, but, if possible, let
G be the centre, and let
GA,
GD,
GB be joined.
Then, since
AD is equal to
DB, and
DG is common,
the two sides AD, DG are equal to the two sides BD, DG respectively; and the base
GA is equal to the base
GB, for they are
radii;
therefore the angle ADG is equal to the angle GDB. [I. 8]
But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [
I. Def. 10]
therefore the angle GDB is right.
But the angle
FDB is also right; therefore the angle
FDB is equal to the angle
GDB, the greater to the less: which is impossible.
Therefore G is not the centre of the circle ABC.
Similarly we can prove that neither is any other point except
F.
Therefore the point F is the centre of the circle ABC.
PORISM.
From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at
right angles, the centre of the circle is on the cutting straight line. Q. E. F.
1
2
PROPOSITION 2.
If on the circumference of a circle two points be taken at random,
the straight line joining the points will fall within the circle.
Let
ABC be a circle, and let two points
A,
B be taken at random on its circumference; I say that the straight line joined from
A to
B will fall within the circle.
For suppose it does not, but, if possible, let it fall outside, as
AEB; let the centre of the circle
ABC be taken [
III. 1], and let it be
D; let
DA,
DB be joined, and let
DFE be drawn through.
Then, since
DA is equal to
DB,
the angle DAE is also equal to the angle DBE. [I. 5] And, since one side
AEB of the triangle
DAE is produced,
the angle DEB is greater than the angle DAE. [I. 16]
But the angle
DAE is equal to the angle
DBE;
therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [
I. 19]
therefore DB is greater than DE. But DB is equal to DF;
therefore DF is greater than DE,
the less than the greater : which is impossible.
Therefore the straight line joined from
A to
B will not fall outside the circle.
Similarly we can prove that neither will it fall on the circumference itself;
therefore it will fall within.
Therefore etc. Q. E. D.
PROPOSITION 3.
If in a circle a straight line through the centre bisect a straight line not through the centre,
it also cuts it at right angles; and if it cut it at right angles,
it also bisects it.
Let
ABC be a circle, and in it let a straight line
CD
through the centre bisect a straight line
AB not through the centre at the point
F; I say that it also cuts it at right angles.
For let the centre of the circle
ABC
be taken, and let it be
E; let
EA,
EB be joined.
Then, since
AF is equal to
FB, and
FE is common,
two sides are equal to two sides;
and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8]
But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [
I. Def. 10]
therefore each of the angles AFE, BFE is right.
Therefore
CD, which is through the centre, and bisects
AB which is not through the centre, also cuts it at right angles.
Again, let
CD cut
AB at right angles;
I say that it also bisects it. that is, that
AF is equal to
FB.
For, with the same construction,
since EA is equal to EB, the angle
EAF is also equal to the angle
EBF. [
I. 5]
But the right angle
AFE is equal to the right angle
BFE,
therefore
EAF,
EBF are two triangles having two angles equal to two angles and one side equal to one side, namely
EF, which is common to them, and subtends one of the equal angles;
therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB.
Therefore etc. Q. E. D.
3
PROPOSITION 4.
If in a circle two straight lines cut one another which are not through the centre,
they do not bisect one another.
Let
ABCD be a circle, and in it let the two straight lines
AC,
BD, which are not through the centre, cut one another at
E; I say that they do not bisect one another.
For, if possible, let them bisect one another, so that
AE is equal to
EC, and
BE to
ED; let the centre of the circle
ABCD be taken [
III. 1], and let it be
F; let
FE be joined.
Then, since a straight line
FE through the centre bisects a straight line
AC not through the centre,
it also cuts it at right angles; [III. 3] therefore the angle FEA is right.
Again, since a straight line
FE bisects a straight line
BD,
it also cuts it at right angles; [III. 3] therefore the angle FEB is right.
But the angle
FEA was also proved right;
therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible.
Therefore
AC,
BD do not bisect one another.
Therefore etc. Q. E. D.
PROPOSITION 5.
If two circles cut one another,
they will not have the same centre.
For let the circles
ABC,
CDG cut one another at the points
B,
C; I say that they will not have the same centre.
For, if possible, let it be
E; let
EC be joined, and let
EFG be drawn through at random.
Then, since the point
E is the centre of the circle
ABC,
EC is equal to EF. [I. Def. 15]
Again, since the point
E is the centre of the circle
CDG,
EC is equal to EG.
But
EC was proved equal to
EF also;
therefore EF is also equal to EG, the less to the greater : which is impossible.
Therefore the point
E is not the centre of the circles
ABC,
CDG.
Therefore etc. Q. E. D.
PROPOSITION 6.
If two circles touch one another,
they will not have the same centre.
For let the two circles
ABC,
CDE touch one another at the point
C; I say that they will not have the same centre.
For, if possible, let it be
F; let
FC be joined, and let
FEB be drawn through at random.
Then, since the point
F is the centre of the circle
ABC,
FC is equal to FB.
Again, since the point
F is the centre of the circle
CDE,
FC is equal to FE.
But
FC was proved equal to
FB;
therefore FE is also equal to FB, the less to the greater: which is impossible.
Therefore
F is not the centre of the circles
ABC,
CDE.
Therefore etc. Q. E. D.
PROPOSITION 7.
If on the diameter of a circle a point be taken which is not the centre of the circle,
and from the point straight lines fall upon the circle,
that will be greatest on which the centre is,
the remainder of the same diameter will be least,
and of the rest
the nearer to the straight line through the centre is always greater than the more remote,
and only two equal straight lines will fall from the point on the circle,
one on each side of the least straight line.
Let
ABCD be a circle, and let
AD be a diameter of it;
on
AD let a point
F be taken which is not the centre of the circle, let
E be the centre of the circle, and from
F let straight lines
FB,
FC,
FG fall upon the circle
ABCD; I say that
FA is greatest,
FD is least, and of the rest
FB is
greater than
FC, and
FC than
FG.
For let
BE,
CE,
GE be joined.
Then, since in any triangle two sides are greater than the remaining one, [
I. 20]
EB, EF are greater than BF.
But
AE is equal to
BE;
therefore AF is greater than BF.
Again, since
BE is equal to
CE, and
FE is common,
the two sides
BE,
EF are equal to the two sides
CE,
EF.
But the angle
BEF is also greater than the angle
CEF; therefore the base
BF is greater than the base
CF. [
I. 24]
For the same reason
CF is also greater than FG.
Again, since
GF,
FE are greater than
EG, and
EG is equal to
ED,
GF, FE are greater than ED.
Let
EF be subtracted from each;
therefore the remainder GF is greater than the remainder FD.
Therefore
FA is greatest,
FD is least, and
FB is greater than
FC, and
FC than
FG.
I say also that from the point
F only two equal straight lines will fall on the circle
ABCD, one on each side of the
least
FD.
For on the straight line
EF, and at the point
E on it, let the angle
FEH be constructed equal to the angle
GEF [
I. 23], and let
FH be joined.
Then, since
GE is equal to
EH,
and
EF is common,
the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF;
therefore the base FG is equal to the base FH. [I. 4]
I say again that another straight line equal to
FG will not
fall on the circle from the point
F.
For, if possible, let
FK so fall.
Then, since
FK is equal to
FG, and
FH to
FG,
FK is also equal to FH, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.
Therefore another straight line equal to
GF will not fall from the point
F upon the circle;
therefore only one straight line will so fall.
Therefore etc. Q. E. D.
4
5
PROPOSITION 8.
If a point be taken outside a circle and from the point straight lines be drawn through to the circle,
one of which is through the centre and the others are drawn at random,
then,
of the straight lines which fall on the concave circumference,
that through the centre is greatest,
while of the rest
the nearer to that through the centre is always greater than the more remote,
but,
of the straight lines falling on the convex circumference,
that between the point and the diameter is least,
while of the rest the nearer to the least is always less than the more remote,
and only two equal straight lines will fall on the circle from the point,
one on each side of the least.
Let
ABC be a circle, and let a point
D be taken outside
ABC; let there be drawn through from it straight lines
DA,
DE,
DF,
DC, and let
DA be through the centre; I say that, of the straight lines falling on the concave circumference
AEFC, the straight line
DA through the centre is greatest, while
DE is greater than
DF and
DF than
DC.; but, of the straight lines falling on the convex circumference
HLKG, the straight line
DG between the point and the diameter
AG is least; and the nearer to the least
DG is always less than the more remote, namely
DK than
DL, and
DL than
DH.
For let the centre of the circle
ABC be taken [
III. 1], and let it be
M; let
ME,
MF,
MC,
MK,
ML,
MH be joined.
Then, since
AM is equal to
EM, let
MD be added to each;
therefore AD is equal to EM, MD.
But
EM,
MD are greater than
ED; [
I. 20]
therefore AD is also greater than ED.
Again, since
ME is equal to
MF,
and MD is common, therefore
EM,
MD are equal to
FM,
MD;
and the angle EMD is greater than the angle FMD;
therefore the base ED is greater than the base FD. [I. 24]
Similarly we can prove that
FD is greater than
CD; therefore
DA is greatest, while
DE is greater than
DF, and
DF than
DC.
Next, since
MK,
KD are greater than
MD, [
I. 20] and
MG is equal to
MK, therefore the remainder
KD is greater than the remainder
GD,
so that GD is less than KD.
And, since on
MD, one of the sides of the triangle
MLD, two straight lines
MK,
KD were constructed meeting within the triangle, therefore
MK,
KD are less than
ML,
LD; [
I. 21] and
MK is equal to
ML;
therefore the remainder DK is less than the remainder DL.
Similarly we can prove that
DL is also less than
DH;
therefore DG is least, while DK is less than DL, and DL than DH.
I say also that only two equal straight lines will fall from the point
D on the circle, one on each side of the least
DG.
On the straight line
MD, and at the point
M on it, let the angle
DMB be constructed equal to the angle
KMD, and let
DB be joined.
Then, since
MK is equal to
MB, and
MD is common,
the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle
KMD is equal to the angle
BMD;
therefore the base DK is equal to the base DB. [I. 4]
I say that no other straight line equal to the straight line
DK will fall on the circle from the point
D.
For, if possible, let a straight line so fall, and let it be
DN.
Then, since DK is equal to DN,
while
DK is equal to
DB,
DB is also equal to DN, that is, the nearer to the least
DG equal to the more remote: which was proved impossible.
Therefore no more than two equal straight lines will fall on the circle
ABC from the point
D, one on each side of
DG the least.
Therefore etc.
PROPOSITION 9.
If a point be taken within a circle,
and more than two equal straight lines fall from the point on the circle,
the point taken is the centre of the circle.
Let
ABC be a circle and
D a point within it, and from
D let more than two equal straight lines, namely
DA,
DB,
DC, fall on the circle
ABC; I say that the point
D is the centre of the circle
ABC.
For let
AB,
BC be joined and bisected at the points
E,
F, and let
ED,
FD be joined and drawn through to the points
G,
K,
H,
L.
Then, since
AE is equal to
EB, and
ED is common,
the two sides AE, ED are equal to the two sides BE, ED; and the base
DA is equal to the base
DB;
therefore the angle AED is equal to the angle BED. [I. 8]
Therefore each of the angles
AED,
BED is right; [
I. Def. 10] therefore
GK cuts
AB into two equal parts and at right angles.
And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [
