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, [
III. 1, Por.]
the centre of the circle is on GK.
For the same reason
the centre of the circle ABC is also on HL.
And the straight lines
GK,
HL have no other point common but the point
D;
therefore the point D is the centre of the circle ABC.
Therefore etc. Q. E. D.
PROPOSITION 10.
A circle does not cut a circle at more points than two.
For, if possible, let the circle
ABC cut the circle
DEF at more points than two, namely
B,
C,
F,
H;
let
BH,
BG be joined and bisected at the points
K,
L, and from
K,
L let
KC,
LM be drawn at right angles to
BH,
BG and carried through to the points
A,
E.
Then, since in the circle
ABC a straight line
AC cuts a straight line
BH into two equal parts and at right angles,
the centre of the circle ABC is on AC. [III. 1, Por.]
Again, since in the same circle
ABC a straight line
NO cuts a straight line
BG into two equal parts and at right angles,
the centre of the circle ABC is on NO.
But it was also proved to be on
AC, and the straight lines
AC,
NO meet at no point except at
P;
therefore the point P is the centre of the circle ABC.
Similarly we can prove that
P is also the centre of the circle
DEF;
therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5]
Therefore etc. Q. E. D.
6
PROPOSITION 11.
If two circles touch one another internally,
and their centres be taken,
the straight line joining their centres,
if it be also produced,
will fall on the point of contact of the circles.
For let the two circles
ABC,
ADE touch one another internally at the point
A, and let the centre
F of the circle
ABC, and the centre
G of
ADE, be taken; I say that the straight line joined from
G to
F and produced will fall on
A.
For suppose it does not, but, if possible, let it fall as
FGH, and let
AF,
AG be joined.
Then, since
AG,
GF are greater than
FA, that is, than
FH,
let
FG be subtracted from each; therefore the remainder
AG is greater than the remainder
GH.
But
AG is equal to
GD;
therefore GD is also greater than GH, the less than the greater: which is impossible.
Therefore the straight line joined from
F to
G will not fall outside;
therefore it will fall at A on the point of contact.
Therefore etc. Q. E. D.
7
8
PROPOSITION 12.
If two circles touch one another externally,
the straight line joining their centres will pass through the point of contact.
For let the two circles
ABC,
ADE touch one another
externally at the point
A, and let the centre
F of
ABC, and the centre
G of
ADE, be taken; I say that the straight line joined from
F to
G will pass through the point of contact at
A.
For suppose it does not,
but, if possible, let it pass as
FCDG, and let
AF,
AG be joined.
Then, since the point
F is the centre of the circle
ABC,
FA is equal to FC.
Again, since the point
G is the centre of the circle
ADE,
GA is equal to GD.
But
FA was also proved equal to
FC;
therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [
I. 20]: which is impossible.
Therefore the straight line joined from
F to
G will not fail to pass through the point of contact at
A;
therefore it will pass through it.
Therefore etc. Q. E. D.]
9
PROPOSITION 13.
A circle does not touch a circle at more points than one,
whether it touch it internally or externally.
For, if possible, let the circle
ABDC touch the circle
EBFD, first internally, at more
points than one, namely
D,
B.
Let the centre
G of the circle
ABDC, and the centre
H of
EBFD, be taken.
Therefore the straight line
joined from
G to
H will fall on
B,
D. [
III. 11]
Let it so fall, as
BGHD.
Then, since the point
G is the centre of the circle
ABCD,
BG is equal to GD;
therefore BG is greater than HD; therefore BH is much greater than HD.
Again, since the point
H is the centre of the circle
EBFD,
BH is equal to HD;
but it was also proved much greater than it: which is impossible.
Therefore a circle does not touch a circle internally at more points than one.
I say further that neither does it so touch it externally.
For, if possible, let the circle
ACK touch the circle
ABDC at more points than one, namely
A,
C, and let
AC be joined.
Then, since on the circumference of each of the circles
ABDC,
ACK two points
A,
C have been taken at random, the straight line joining the points will fall within each circle; [
III. 2]
but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd.
Therefore a circle does not touch a circle externally at more points than one.
And it was proved that neither does it so touch it internally.
Therefore etc. Q. E. D.
10
11
PROPOSITION 14.
In a circle equal straight lines are equally distant from the centre,
and those which are equally distant from the centre are equal to one another.
Let
ABDC be a circle, and let
AB,
CD be equal straight lines in it; I say that
AB,
CD are equally distant from the centre.
For let the centre of the circle
ABDC be taken [
III. 1], and let it be
E; from
E let
EF,
EG be drawn perpendicular to
AB,
CD, and let
AE,
EC be joined.
Then, since a straight line
EF through the centre cuts a straight line
AB not through the centre at right angles, it also bisects it. [
III. 3]
Therefore AF is equal to FB; therefore AB is double of AF.
For the same reason
CD is also double of CG; and
AB is equal to
CD;
therefore AF is also equal to CG.
And, since
AE is equal to
EC,
the square on AE is also equal to the square on EC. But the squares on
AF,
EF are equal to the square on
AE, for the angle at
F is right; and the squares on
EG,
GC are equal to the square on
EC, for the angle at
G is right; [
I. 47]
therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on
AF is equal to the square on
CG, for
AF is equal to
CG;
therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG.
But 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; [
III. Def. 4]
therefore AB, CD are equally distant from the centre.
Next, let the straight lines
AB,
CD be equally distant from the centre; that is, let
EF be equal to
EG.
I say that
AB is also equal to
CD.
For, with the same construction, we can prove, similarly, that
AB is double of
AF, and
CD of
CG.
And, since
AE is equal to
CE,
the square on AE is equal to the square on CE. But the squares on
EF,
FA are equal to the square on
AE, and the squares on
EG,
GC equal to the square on
CE. [
I. 47]
Therefore the squares on
EF,
FA are equal to the squares on
EG,
GC, of which the square on
EF is equal to the square on
EG, for
EF is equal to
EG; therefore the square on
AF which remains is equal to the square on
CG;
therefore AF is equal to CG. And
AB is double of
AF, and
CD double of
CG;
therefore AB is equal to CD.
Therefore etc. Q. E. D.
PROPOSITION 15.
Of straight lines in a circle the diameter is greatest,
and of the rest the nearer to the centre is always greater than the more remote.
Let
ABCD be a circle, let
AD be its diameter and
E the centre; and let
BC be nearer to the diameter
AD, and
FG more remote; I say that
AD is greatest and
BC greater than
FG.
For from the centre
E let
EH,
EK
be drawn perpendicular to
BC,
FG.
Then, since
BC is nearer to the centre and
FG more remote,
EK is greater than
EH. [
III. Def. 5]
Let
EL be made equal to
EH, through
L let
LM be drawn at right angles to
EK and carried through to
N, and let
ME,
EN,
FE,
EG be joined.
Then, since
EH is equal to
EL,
BC is also equal to MN. [III. 14]
Again, since
AE is equal to
EM, and
ED to
EN,
AD is equal to ME, EN.
But
ME,
EN are greater than
MN, [
I. 20] and
MN is equal to
BC;
therefore AD is greater than BC.
And, since the two sides
ME,
EN are equal to the two sides
FE,
EG, and the angle
MEN greater than the angle
FEG, therefore the base
MN is greater than the base
FG. [
I. 24]
But
MN was proved equal to
BC.
Therefore the diameter
AD is greatest and
BC greater than
FG.
Therefore etc. Q. E. D.
12
13
PROPOSITION 16.
The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle,
and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater,
and the remaining angle less,
than any acute rectilineal angle.
Let
ABC be a circle about
D as centre and
AB as diameter; I say that the straight line drawn from
A at right angles to
AB from its extremity will fall outside the circle.
For suppose it does not, but, if possible, let it fall within as
CA, and let
DC be joined.
Since
DA is equal to
DC,
the angle DAC is also equal to the angle ACD. [I. 5]
But the angle
DAC is right;
therefore the angle ACD is also right: thus, in the triangle
ACD, the two angles
DAC,
ACD are equal to two right angles: which is impossible. [
I. 17]
Therefore the straight line drawn from the point
A at right angles to
BA will not fall within the circle.
Similarly we can prove that neither will it fall on the circumference;
therefore it will fall outside.
Let it fall as
AE; I say next that into the space between the straight line
AE and the circumference
CHA another straight line cannot be interposed.
For, if possible, let another straight line be so interposed, as
FA, and let
DG be drawn from the point
D perpendicular to
FA.
Then, since the angle
AGD is right,
and the angle DAG is less than a right angle, AD is greater than DG. [I. 19]
But
DA is equal to
DH;
therefore DH is greater than DG, the less than the greater: which is impossible.
Therefore another straight line cannot be interposed into the space between the straight line and the circumference.
I say further that the angle of the semicircle contained by the straight line
BA and the circumference
CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference
CHA and the straight line
AE is less than any acute rectilineal angle.
For, if there is any rectilineal angle greater than the angle contained by the straight line
BA and the circumference
CHA, and any rectilineal angle less than the angle contained by the circumference
CHA and the straight line
AE, then into the space between the circumference and the straight line
AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line
BA and the circumference
CHA, and another angle contained by straight lines which is less than the angle contained by the circumference
CHA and the straight line
AE.
But such a straight line cannot be interposed;
therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line
BA and the circumference
CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference
CHA and the straight line
AE.—
PORISM.
From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its extremity touches the circle. Q. E. D.
14
PROPOSITION 17.
From a given point to draw a straight line touching a given circle.
Let
A be the given point, and
BCD the given circle; thus it is required to draw from the point
A a straight line touching the circle
BCD.
For let the centre
E of the circle be taken; [
III. 1] let
AE be joined, and with centre
E and distance
EA let the circle
AFG be described; from
D let
DF be drawn at right angles to
EA, and let
EF,
AB be joined; I say that
AB has been drawn from the point
A touching the circle
BCD.
For, since
E is the centre of the circles
BCD,
AFG,
EA is equal to EF, and ED to EB; therefore the two sides
AE,
EB are equal to the two sides
FE,
ED: and they contain a common angle, the angle at
E;
therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [I. 4] therefore the angle EDF is equal to the angle EBA.
But the angle
EDF is right;
therefore the angle EBA is also right.
Now
EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [
III. 16, Por.]
therefore AB touches the circle BCD.
Therefore from the given point
A the straight line
AB has been drawn touching the circle
BCD.
PROPOSITION 18.
If a straight line touch a circle,
and a straight line be joined from the centre to the point of contact,
the straight line so joined will be perpendicular to the tangent.
For let a straight line
DE touch the circle
ABC at the point
C, let the centre
F of the circle
ABC be taken, and let
FC be joined from
F to
C; I say that
FC is perpendicular to
DE.
For, if not, let
FG be drawn from
F perpendicular to
DE.
Then, since the angle
FGC is right,
the angle FCG is acute; [I. 17] and the greater angle is subtended by the greater side; [
I. 19]
therefore FC is greater than FG.
But
FC is equal to
FB;
therefore FB is also greater than FG, the less than the greater: which is impossible.
Therefore
FG is not perpendicular to
DE.
Similarly we can prove that neither is any other straight line except
FC;
therefore FC is perpendicular to DE.
Therefore etc. Q. E. D.
15
PROPOSITION 19.
If a straight line touch a circle,
and from the point of contact a straight line be drawn at right angles to the tangent,
the centre of the circle will be on the straight line so drawn.
For let a straight line
DE touch the circle
ABC at the point
C, and from
C let
CA be drawn at right angles to
DE; I say that the centre of the circle is on
AC.
For suppose it is not, but, if possible, let
F be the centre, and let
CF be joined.
Since a straight line
DE touches the circle
ABC, and
FC has been joined from the centre to the point of contact,
FC is perpendicular to DE; [III. 18] therefore the angle FCE is right.
But the angle
ACE is also right;
therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible.
Therefore
F is not the centre of the circle
ABC.
Similarly we can prove that neither is any other point except a point on
AC.
Therefore etc. Q. E. D.
PROPOSITION 20.
In a circle the angle at the centre is double of the angle at the circumference,
when the angles have the same circumference as base.
Let
ABC be a circle, let the angle
BEC be an angle
at its centre, and the angle
BAC an angle at the circumference, and let them have the same circumference
BC as base; I say that the angle
BEC is double of
the angle
BAC.
For let
AE be joined and drawn through to
F.
Then, since
EA is equal to
EB,
the angle EAB is also equal to the angle EBA; [I. 5] therefore the angles
EAB,
EBA are double of the angle
EAB.
But the angle
BEF is equal to the angles
EAB,
EBA; [
I. 32] therefore the angle
BEF is also double of the angle
EAB.
For the same reason
the angle FEC is also double of the angle EAC.
Therefore the whole angle
BEC is double of the whole angle
BAC.
Again let another straight line be inflected, and let there be another angle
BDC; let
DE be joined and produced to
G.
Similarly then we can prove that the angle
GEC is double of the angle
EDC,
of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC.
Therefore etc. Q. E. D.
17
PROPOSITION 21.
In a circle the angles in the same segment are equal to one another.
Let
ABCD be a circle, and let the angles
BAD,
BED be angles in the same segment
BAED; I say that the angles
BAD,
BED are equal to one another.
For let the centre of the circle
ABCD be taken, and let it be
F; let
BF,
FD be joined.
Now, since the angle
BFD is at the centre,
and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double of the angle BAD. [III. 20]
For the same reason
the angle BFD is also double of the angle BED; therefore the angle BAD is equal to the angle BED.
Therefore etc. Q. E. D.
PROPOSITION 22.
The opposite angles of quadrilaterals in circles are equal to two right angles.
Let
ABCD be a circle, and let
ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles.
Let
AC,
BD be joined.
Then, since in any triangle the three angles are equal to two right angles, [
I. 32]
the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles.
But the angle
CAB is equal to the angle
BDC, for they are in the same segment
BADC; [
III. 21] and the angle
ACB is equal to the angle
ADB, for they are in the same segment
ADCB; therefore the whole angle
ADC is equal to the angles
BAC,
ACB.
Let the angle
ABC be added to each; therefore the angles
ABC,
BAC,
ACB are equal to the angles
ABC,
ADC. But the angles
ABC,
BAC,
ACB are equal to two right angles; therefore the angles
ABC,
ADC are also equal to two right angles.
Similarly we can prove that the angles
BAD,
DCB are also equal to two right angles.
Therefore etc. Q. E. D.
PROPOSITION 23.
On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side.
For, if possible, on the same straight line
AB let two similar and unequal segments of circles
ACB,
ADB be constructed on the same side; let
ACD be drawn through, and let
CB,
DB be joined.
Then, since the segment
ACB is similar to the segment
ADB, and similar segments of circles are those which admit equal angles, [
III. Def. 11]
the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16]
Therefore etc. Q. E. D.
18
PROPOSITION 24.
Similar segments of circles on equal straight lines are equal to one another.
For let
AEB,
CFD be similar segments of circles on equal straight lines
AB,
CD;
I say that the segment
AEB is equal to the segment
CFD.
For, if the segment
AEB be applied to
CFD, and if the point
A be placed on
C and the straight line
AB on
CD,
the point B will also coincide with the point D, because AB is equal to CD;
and, AB coinciding with CD, the segment AEB will also coincide with CFD.
For, if the straight line
AB coincide with
CD but the segment
AEB do not coincide with
CFD, it will either fall with it, or outside it;
or it will fall awry, as
CGD, and a circle cuts a circle at more points than two : which is impossible. [
III. 10]
Therefore, if the straight line
AB be applied to
CD, the segment
AEB will not fail to coincide with
CFD also;
therefore it will coincide with it and will be equal to it.
Therefore etc. Q. E. D.
19
PROPOSITION 25.
Given a segment of a circle,
to describe the complete circle of which it is a segment.
Let
ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment
ABC, that is, of which it is a segment.
For let
AC be bisected at
D, let
DB be drawn from the point
D at right angles to
AC, and let
AB. be joined;
the angle ABD is then greater than, equal to, or less than the angle BAD.
First let it be greater; and on the straight line
BA, and at the point
A on it, let the angle
BAE be constructed equal to the angle
ABD; let
DB be drawn through to
E, and let
EC be joined.
Then, since the angle
ABE is equal to the angle
BAE,
the straight line EB is also equal to EA. [I. 6]
And, since
AD is equal to
DC, and
DE is common,
the two sides AD, DE are equal to the two sides CD, DE respectively; and the angle
ADE is equal to the angle
CDE, for each is right;
therefore the base AE is equal to the base CE.
But
AE was proved equal to
BE;
therefore BE is also equal to CE; therefore the three straight lines
AE,
EB,
EC are equal to one another.
Therefore the circle drawn with centre
E and distance one of the straight lines
AE,
EB,
EC will also pass through the remaining points and will have been completed. [
III. 9]
Therefore, given a segment of a circle, the complete circle has been described.
And it is manifest that the segment
ABC is less than a semicircle, because the centre
E happens to be outside it.
Similarly, even if the angle
ABD be equal to the angle
BAD,
AD being equal to each of the two
BD,
DC,
the three straight lines DA, DB, DC will be equal to one another,
D will be the centre of the completed circle, and ABC will clearly be a semicircle.
But, if the angle
ABD be less than the angle
BAD, and if we construct, on the straight line
BA and at the point
A on it, an angle equal to the angle
ABD, the centre will fall on
DB within the segment
ABC, and the segment
ABC will clearly be greater than a semicircle.
Therefore, given a segment of a circle, the complete circle has been described. Q. E. F.
20
PROPOSITION 26.
In equal circles equal angles stand on equal circumferences,
whether they stand at the centres or at the circumferences.
Let
ABC,
DEF be equal circles, and in them let there be equal angles, namely at the centres the angles
BGC,
EHF, and at the circumferences the angles
BAC,
EDF; I say that the circumference
BKC is equal to the circumference
ELF.
For let
BC,
EF be joined.
Now, since the circles
ABC,
DEF are equal,
the radii are equal.
Thus the two straight lines
BG,
GC are equal to the two straight lines
EH,
HF;
and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. [I. 4]
And, since the angle at
A is equal to the angle at
D,
the segment BAC is similar to the segment EDF; [III. Def. 11] and they are upon equal straight lines.
But similar segments of circles on equal straight lines are equal to one another; [
III. 24]
therefore the segment BAC is equal to EDF. But the whole circle
ABC is also equal to the whole circle
DEF; therefore the circumference
BKC which remains is equal to the circumference
ELF.
Therefore etc. Q. E. D.
PROPOSITION 27.
In equal circles angles standing on equal circumferences are equal to one another,
whether they stand at the centres or at the circumferences.
For in equal circles
ABC,
DEF, on equal circumferences
BC,
EF, let the angles
BGC,
EHF stand at the centres
G,
H, and the angles
BAC,
EDF at the circumferences; I say that the angle
BGC is equal to the angle
EHF, and the angle
BAC is equal to the angle
EDF.
For, if the angle
BGC is unequal to the angle
EHF,
one of them is greater. Let the angle
BGC be greater : and on the straight line
BG, and at the point
G on it, let the angle
BGK be constructed equal to the angle
EHF. [
I. 23]
Now equal angles stand on equal circumferences, when they are at the centres; [
III. 26]
therefore the circumference BK is equal to the circumference EF.
But
EF is equal to
BC;
therefore BK is also equal to BC, the less to the greater : which is impossible.
Therefore the angle
BGC is not unequal to the angle
EHF;
therefore it is equal to it.
And the angle at
A is half of the angle
BGC,
and the angle at D half of the angle EHF; [III. 20] therefore the angle at
A is also equal to the angle at
D.
Therefore etc. Q. E. D.
PROPOSITION 28.
In equal circles equal straight lines cut off equal circumferences,
the greater equal to the greater and the less to the less.
Let
ABC,
DEF be equal circles, and in the circles let
AB,
DE be equal straight lines cutting off
ACB,
DFE as greater circumferences and
AGB,
DHE as lesser; I say that the greater circumference
ACB is equal to the greater circumference
DFE, and the less circumference
AGB to
DHE.
For let the centres
K,
L of the circles be taken, and let
AK,
KB,
DL,
LE be joined.
Now, since the circles are equal,
the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base
AB is equal to the base
DE;
therefore the angle AKB is equal to the angle DLE. [I. 8]
But equal angles stand on equal circumferences, when they are at the centres; [
III. 26]
therefore the circumference AGB is equal to DHE.
And the whole circle
ABC is also equal to the whole circle
DEF; therefore the circumference
ACB which remains is also equal to the circumference
DFE which remains.
Therefore etc. Q. E. D.
PROPOSITION 29.
In equal circles equal circumferences are subtended by equal straight lines.
Let
ABC,
DEF be equal circles, and in them let equal circumferences
BGC,
EHF be cut off; and let the straight lines
BC,
EF be joined; I say that
BC is equal to
EF.
For let the centres of the circles be taken, and let them be
K,
L; let
BK,
KC,
EL,
LF be joined.
Now, since the circumference
BGC is equal to the circumference
EHF,
the angle BKC is also equal to the angle ELF. [III. 27]
And, since the circles
ABC,
DEF are equal,
the radii are also equal; therefore the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I. 4]
Therefore etc.
PROPOSITION 30.
To bisect a given circumference.
Let
ADB be the given circumference; thus it is required to bisect the circumference
ADB.
Let
AB be joined and bisected at
C; from the point
C let
CD be drawn at right angles to the straight line
AB, and let
AD,
DB be joined.
Then, since
AC is equal to
CB, and
CD is common,
the two sides AC, CD are equal to the two sides BC, CD; and the angle
ACD is equal to the angle
BCD, for each is right;
therefore the base AD is equal to the base DB. [I. 4]
But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [
III. 28]
and each of the circumferences AD, DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB.
Therefore the given circumference has been bisected at the point
D. Q. E. F.
PROPOSITION 31.
In a circle the angle in the semicircle is right,
that in a greater segment less than a right angle,
and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle,
and the angle of the less segment less than a right angle.
Let
ABCD be a circle, let
BC be its diameter, and
E its centre, and let
BA,
AC,
AD,
DC be joined; I say that the angle
BAC in the semicircle
BAC is right, the angle
ABC in the segment
ABC greater than the semicircle is less than a right angle, and the angle
ADC in the segment
ADC less than the semicircle is greater than a right angle.
Let
AE be joined, and let
BA be carried through to
F.
Then, since
BE is equal to
EA,
the angle ABE is also equal to the angle BAE. [I. 5]
Again, since
CE is equal to
EA,
the angle ACE is also equal to the angle CAE. [I. 5]
Therefore the whole angle
BAC is equal to the two angles
ABC,
ACB.
But the angle
FAC exterior to the triangle
ABC is also equal to the two angles
ABC,
ACB; [
I. 32]
therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10] therefore the angle BAC in the semicircle BAC is right.
Next, since in the triangle
ABC the two angles
ABC,
BAC are less than two right angles, [
I. 17] and the angle
BAC is a right angle,
the angle ABC is less than a right angle; and it is the angle in the segment
ABC greater than the semicircle.
Next, since
ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [
III. 22] while the angle
ABC is less than a right angle, therefore the angle
ADC which remains is greater than a right angle; and it is the angle in the segment
ADC less than the semicircle.
I say further that the angle of the greater segment, namely that contained by the circumference
ABC and the straight line
AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference
ADC and the straight line
AC, is less than a right angle.
This is at once manifest. For, since the angle contained by the straight lines
BA,
AC is right,
the angle contained by the circumference ABC and the straight line AC is greater than a right angle.
Again, since the angle contained by the straight lines
AC,
AF is right,
the angle contained by the straight line CA and the circumference ADC is less than a right angle.
Therefore etc. Q. E. D.
PROPOSITION 32.
If a straight line touch a circle,
and from the point of contact there be drawn across,
in the circle,
a straight line cutting the circle,
the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.
For let a straight line
EF touch the circle
ABCD at the point
B, and from the point
B let there be drawn across, in the circle
ABCD, a straight line
BD cutting it; I say that the angles which
BD makes with the tangent
EF will be equal to the angles in the alternate segments of the circle, that is, that the angle
FBD is equal to the angle constructed in the segment
BAD, and the angle
EBD is equal to the angle constructed in the segment
DCB.
For let
BA be drawn from
B at right angles to
EF, let a point
C be taken at random on the circumference
BD, and let
AD,
DC,
CB be joined.
Then, since a straight line
EF touches the circle
ABCD at
B, and
BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle
ABCD is on
BA. [
III. 19]
Therefore
BA is a diameter of the circle
ABCD;
therefore the angle ADB, being an angle in a semicircle, is right. [III. 31]
Therefore the remaining angles
BAD,
ABD are equal to one right angle. [
I. 32]
But the angle
ABF is also right; therefore the angle
ABF is equal to the angles
BAD,
ABD.
Let the angle
ABD be subtracted from each; therefore the angle
DBF which remains is equal to the angle
BAD in the alternate segment of the circle.
Next, since
ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [
III. 22]
But the angles
DBF,
DBE are also equal to two right angles; therefore the angles
DBF,
DBE are equal to the angles
BAD,
BCD,
of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle.
Therefore etc. Q. E. D.
PROPOSITION 33.
On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.
Let
AB be the given straight line, and the angle at
C the given rectilineal angle; thus it is required to describe on the given straight line
AB a segment of a circle admitting an angle equal to the angle at
C.
The angle at
C is then acute, or right, or obtuse.
First let it be acute, and, as in the first figure, on the straight line
AB, and at the point
A, let the angle
BAD be constructed equal to the angle at
C;
therefore the angle BAD is also acute.
Let
AE be drawn at right angles to
DA, let
AB be bisected at
F, let
FG be drawn from the point
F at right angles to
AB, and let
GB be joined.
Then, since
AF is equal to
FB, and
FG is common,
the two sides AF, FG are equal to the two sides BF, FG; and the angle
AFG is equal to the angle
BFG;
therefore the base AG is equal to the base BG. [I. 4]
Therefore the circle described with centre
G and distance
GA will pass through
B also.
Let it be drawn, and let it be
ABE; let
EB be joined.
Now, since
AD is drawn from
A, the extremity of the diameter
AE, at right angles to
AE,
therefore AD touches the circle ABE. [III. 16, Por.]
Since then a straight line
AD touches the circle
ABE, and from the point of contact at
A a straight line
AB is drawn across in the circle
ABE,
the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32]
But the angle
DAB is equal to the angle at
C; therefore the angle at
C is also equal to the angle
AEB.
Therefore on the given straight line
AB the segment
AEB of a circle has been described admitting the angle
AEB equal to the given angle, the angle at
C.
Next let the angle at
C be right; and let it be again required to describe on
AB a segment of a circle admitting an angle equal to the right angle at
C.
Let the angle
BAD be constructed equal to the right angle at
C, as is the case in the second figure; let
AB be bisected at
F, and with centre
F and distance either
FA or
FB let the circle
AEB be described.
Therefore the straight line
AD touches the circle
ABE, because the angle at
A is right. [
III. 16, Por.]
And the angle
BAD is equal to the angle in the segment
AEB, for the latter too is itself a right angle, being an angle in a semicircle. [
III. 31]
But the angle
BAD is also equal to the angle at
C.
Therefore the angle
AEB is also equal to the angle at
C.
Therefore again the segment
AEB of a circle has been described on
AB admitting an angle equal to the angle at
C.
Next, let the angle at
C be obtuse; and on the straight line
AB, and at the point
A, let the angle
BAD be constructed equal to it, as is the case in the third figure; let
AE be drawn at right angles to
AD, let
AB be again bisected at
F, let
FG be drawn at right angles to
AB, and let
GB be joined.
Then, since
AF is again equal to
FB, and
FG is common,
the two sides AF, FG are equal to the two sides BF, FG; and the angle
AFG is equal to the angle
BFG;
therefore the base AG is equal to the base BG. [I. 4]
Therefore the circle described with centre
G and distance
GA will pass through
B also; let it so pass, as
AEB.
Now, since
AD is drawn at right angles to the diameter
AE from its extremity,
AD touches the circle AEB. [III. 16, Por.]
And
AB has been drawn across from the point of contact at
A;
therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32]
But the angle
BAD is equal to the angle at
C.
Therefore the angle in the segment
AHB is also equal to the angle at
C.
Therefore on the given straight line
AB the segment
AHB of a circle has been described admitting an angle equal to the angle at
C. Q. E. F.
PROPOSITION 34.
From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.
Let
ABC be the given circle, and the angle at
D the given rectilineal angle; thus it is required to cut off from the circle
ABC a segment admitting an angle equal to the given rectilineal angle, the angle at
D.
Let
EF be drawn touching
ABC at the point
B, and on the straight line
FB, and at the point
B on it, let the angle
FBC be constructed equal to the angle at
D. [
I. 23]
Then, since a straight line
EF touches the circle
ABC,
and BC has been drawn across from the point of contact at B, the angle
FBC is equal to the angle constructed in the alternate segment
BAC. [
III. 32]
But the angle
FBC is equal to the angle at
D;
therefore the angle in the segment BAC is equal to the angle at D.
Therefore from the given circle
ABC the segment
BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at
D. Q. E. F.
PROPOSITION 35.
If in a circle two straight lines cut one another,
the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.
For in the circle
ABCD let the two straight lines
AC,
BD cut one another at the point
E; I say that the rectangle contained by
AE,
EC is equal to the rectangle contained by
DE,
EB.
If now
AC,
BD are through the centre, so that
E is the centre of the circle
ABCD, it is manifest that,
AE,
EC,
DE,
EB being equal, the rectangle contained by
AE,
EC is also equal to the rectangle contained by
DE,
EB.
Next let
AC,
DB not be through the centre; let the centre of
ABCD be taken, and let it be
F; from
F let
FG,
FH be drawn perpendicular to the straight lines
AC,
DB, and let
FB,
FC,
FE be joined.
Then, since a straight line
GF through the centre cuts a straight line
AC not through the centre at right angles,
it also bisects it; [III. 3] therefore AG is equal to GC.
Since, then, the straight line
AC has been cut into equal parts at
G and into unequal parts at
E, the rectangle contained by
AE,
EC together with the square on
EG is equal to the square on
GC; [
II. 5]
Let the square on
GF be added; therefore the rectangle
AE,
EC together with the squares on
GE,
GF is equal to the squares on
CG,
GF.
But the square on
FE is equal to the squares on
EG,
GF, and the square on
FC is equal to the squares on
CG,
GF; [
I. 47]
therefore the rectangle AE, EC together with the square on FE is equal to the square on FC.
And
FC is equal to
FB; therefore the rectangle
AE,
EC together with the square on
EF is equal to the square on
FB.
For the same reason, also, the rectangle
DE,
EB together with the square on
FE is equal to the square on
FB.
But the rectangle
AE,
EC together with the square on
FE was also proved equal to the square on
FB; therefore the rectangle
AE,
EC together with the square on
FE is equal to the rectangle
DE,
EB together with the square on
FE.
Let the square on
FE be subtracted from each; therefore the rectangle contained by
AE,
EC which remains is equal to the rectangle contained by
DE,
EB.
Therefore etc.
PROPOSITION 36.
If a point be taken outside a circle and from it there fall on the circle two straight lines,
and if one of them cut the circle and the other touch it,
the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.
For let a point
D be taken outside the circle
ABC, and from
D let the two straight lines
DCA,
DB fall on the circle
ABC; let
DCA cut the circle
ABC and let
BD touch it; I say that the rectangle contained by
AD,
DC is equal to the square on
DB.
Then
DCA is either through the centre or not through the centre.
First let it be through the centre, and let
F be the centre of the circle
ABC; let
FB be joined;
therefore the angle FBD is right. [III. 18]
And, since
AC has been bisected at
F, and
CD is added to it, the rectangle
AD,
DC together with the square on
FC is equal to the square on
FD. [
II. 6]
But
FC is equal to
FB; therefore the rectangle
AD,
DC together with the square on
FB is equal to the square on
FD.
And the squares on
FB,
BD are equal to the square on
FD; [
I. 47] therefore the rectangle
AD,
DC together with the square on
FB is equal to the squares on
FB,
BD.
Let the square on
FB be subtracted from each; therefore the rectangle
AD,
DC which remains is equal to the square on the tangent
DB.
Again, let
DCA not be through the centre of the circle
ABC; let the centre
E be taken, and from
E let
EF be drawn perpendicular to
AC; let
EB,
EC,
ED be joined.
Then the angle
EBD is right. [
III. 18]
And, since a straight line
EF through the centre cuts a straight line
AC not through the centre at right angles,
it also bisects it; [III. 3] therefore AF is equal to FC.
Now, since the straight line
AC has been bisected at the point
F, and
CD is added to it, the rectangle contained by
AD,
DC together with the square on
FC is equal to the square on
FD. [
II. 6]
Let the square on
FE be added to each; therefore the rectangle
AD,
DC together with the squares on
CF,
FE is equal to the squares on
FD,
FE.
But the square on
EC is equal to the squares on
CF,
FE, for the angle
EFC is right; [
I. 47] and the square on
ED is equal to the squares on
DF,
FE; therefore the rectangle
AD,
DC together with the square on
EC is equal to the square on
ED.
And
EC 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, for the angle
EBD is right; [
I. 47] therefore the rectangle
AD,
DC together with the square on
EB is equal to the squares on
EB,
BD.
Let the square on
EB be subtracted from each; therefore the rectangle
AD,
DC which remains is equal to the square on
DB.
Therefore etc. Q. E. D.
PROPOSITION 37.
If a point be taken outside a circle and from the point there fall on the circle two straight lines,
if one of them cut the circle,
and the other fall on it,
and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle,
the straight line which falls on it will touch the circle.
For let a point
D be taken outside the circle
ABC; from
D let the two straight lines
DCA,
DB fall on the circle
ACB; let
DCA cut the circle and
DB fall on it; and let the rectangle
AD,
DC be equal to the square on
DB.
I say that
DB touches the circle
ABC.
For let
DE be drawn touching
ABC; let the centre of the circle
ABC be taken, and let it be
F; let
FE,
FB,
FD be joined.
Thus the angle
FED is right. [
III. 18]
Now, since
DE touches the circle
ABC, and
DCA cuts it, the rectangle
AD,
DC is equal to the square on
DE. [
III. 36]
But the rectangle
AD,
DC was also equal to the square on
DB; therefore the square on
DE is equal to the square on
DB;
therefore DE is equal to DB.
And
FE is equal to
FB; therefore the two sides
DE,
EF are equal to the two sides
DB,
BF; and
FD is the common base of the triangles;
therefore the angle DEF is equal to the angle DBF. [I. 8]
But the angle
DEF is right;
therefore the angle DBF is also right.
And
FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [
III. 16, Por.]
therefore DB touches the circle.
Similarly this can be proved to be the case even if the centre be on
AC.
Therefore etc. Q. E. D.