BOOK IV.
DEFINITIONS.
1
A rectilineal figure is said to be
inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed.
2
Similarly a figure is said to be
circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed.
3
A rectilineal figure is said to be
inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.
4
A rectilineal figure is said to be
circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle.
5
Similarly a circle is said to be
inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.
6
A circle is said to be
circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.
7
A straight line is said to be
fitted into a circle when its extremities are on the circumference of the circle.
BOOK IV. PROPOSITIONS
PROPOSITION 1.
Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle.
Let
ABC be the given circle, and
D the given straight line not greater than the diameter of the circle; thus it is required to fit into the circle
ABC a straight line equal to the straight line
D.
Let a diameter
BC of the circle
ABC be drawn.
Then, if
BC is equal to
D, that which was enjoined will have been done; for
BC has been fitted into the circle
ABC equal to the straight line
D.
But, if
BC is greater than
D, let
CE be made equal to
D, and with centre
C and distance
CE let the circle
EAF be described; let
CA be joined.
Then, since the point
C is the centre of the circle
EAF,
CA is equal to CE.
But
CE is equal to
D;
therefore D is also equal to CA.
Therefore into the given circle
ABC there has been fitted
CA equal to the given straight line
D.
PROPOSITION 2.
In a given circle to inscribe a triangle equiangular with a given triangle.
Let
ABC be the given circle, and
DEF the given triangle; thus it is required to inscribe in the circle
ABC a triangle equiangular with the triangle
DEF.
Let
GH be drawn touching the circle
ABC at
A [
III. 16, Por.]; on the straight line
AH, and at the point
A on it, let the angle
HAC be constructed equal to the angle
DEF, and on the straight line
AG, and at the point
A on it, let the angle
GAB be constructed equal to the angle
DFE; [
I. 23] let
BC be joined.
Then, since a straight line
AH touches the circle
ABC, and from the point of contact at
A the straight line
AC is drawn across in the circle, therefore the angle
HAC is equal to the angle
ABC in the alternate segment of the circle. [
III. 32]
But the angle
HAC is equal to the angle
DEF; therefore the angle
ABC is also equal to the angle
DEF.
For the same reason
the angle ACB is also equal to the angle DFE; therefore the remaining angle
BAC is also equal to the remaining angle
EDF. [
I. 32]
Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle. Q. E. F.
PROPOSITION 3.
About a given circle to circumscribe a triangle equiangular with a given triangle.
Let
ABC be the given circle, and
DEF the given triangle;
thus it is required to circumscribe about the circle
ABC a triangle equiangular with the triangle
DEF.
Let
EF be produced in both directions to the points
G,
H, let the centre
K of the circle
ABC be taken [
III. 1], and let
the straight line
KB be drawn across at random; on the straight line
KB, and at the point
K on it, let the angle
BKA be constructed equal to the angle
DEG, and the angle
BKC equal to the angle
DFH; [
I. 23] and through the points
A,
B,
C let
LAM,
MBN,
NCL be
drawn touching the circle
ABC. [
III. 16, Por.]
Now, since
LM,
MN,
NL touch the circle
ABC at the points
A,
B,
C, and
KA,
KB,
KC have been joined from the centre
K to the points
A,
B,
C,
therefore the angles at the points
A,
B,
C are right. [
III. 18]
And, since the four angles of the quadrilateral
AMBK are equal to four right angles, inasmuch as
AMBK is in fact divisible into two triangles,
and the angles KAM, KBM are right,
therefore the remaining angles
AKB,
AMB are equal to two right angles.
But the angles
DEG,
DEF are also equal to two right angles; [
I. 13] therefore the angles
AKB,
AMB are equal to the angles
DEG,
DEF, of which the angle
AKB is equal to the angle
DEG;
therefore the angle AMB which remains is equal to the angle DEF which remains.
Similarly it can be proved that the angle
LNB is also
equal to the angle
DFE;
therefore the remaining angle MLN is equal to the angle EDF. [I. 32]
Therefore the triangle
LMN is equiangular with the triangle
DEF; and it has been circumscribed about the
circle
ABC.
Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle. Q. E. F.
1
2
PROPOSITION 4.
In a given triangle to inscribe a circle.
Let
ABC be the given triangle; thus it is required to inscribe a circle in the triangle
ABC.
Let the angles
ABC,
ACB
be bisected by the straight lines
BD,
CD [
I. 9], and let these meet one another at the point
D; from
D let
DE,
DF,
DG be drawn perpendicular to the straight
lines
AB,
BC,
CA.
Now, since the angle
ABD is equal to the angle
CBD, and the right angle
BED is also equal to the right angle
BFD,
EBD,
FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is
BD common to the triangles;
therefore they will also have the remaining sides equal to the remaining sides; [I. 26]
therefore DE is equal to DF.
For the same reason
DG is also equal to DF.
Therefore the three straight lines
DE,
DF,
DG are equal
to one another;
therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G
are right.
For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [
III. 16]
therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA;
therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]
Let it be inscribed, as
FGE.
Therefore in the given triangle
ABC the circle
EFG has been inscribed. Q. E. F.
3
PROPOSITION 5.
About a given triangle to circumscribe a circle.
Let
ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle
ABC.
Let the straight lines
AB,
AC be bisected at the points
D,
E [
I. 10], and from the points
D,
E let
DF,
EF be drawn at right angles to
AB,
AC; they will then meet within the triangle
ABC, or on the straight line
BC, or outside
BC.
First let them meet within at
F, and let
FB,
FC,
FA be joined.
Then, since
AD is equal to
DB, and
DF is common and at right angles, therefore the base
AF is equal to the base
FB. [
I. 4]
Similarly we can prove that
CF is also equal to AF; so that
FB is also equal to
FC;
therefore the three straight lines FA, FB, FC are equal to one another.
Therefore the circle described with centre
F and distance one of the straight lines
FA,
FB,
FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle
ABC.
Let it be circumscribed, as
ABC.
Next, let
DF,
EF meet on the straight line
BC at
F, as is the case in the second figure; and let
AF be joined.
Then, similarly, we shall prove that the point
F is the centre of the circle circumscribed about the triangle
ABC.
Again, let
DF,
EF meet outside the triangle
ABC at
F, as is the case in the third figure, and let
AF,
BF,
CF be joined.
Then again, since
AD is equal to
DB, and
DF is common and at right angles, therefore the base
AF is equal to the base
BF. [
I. 4]
Similarly we can prove that
CF is also equal to AF; so that
BF is also equal to
FC; therefore the circle described with centre
F and distance one of the straight lines
FA,
FB,
FC will pass also through the remaining points, and will have been circumscribed about the triangle
ABC.
Therefore about the given triangle a circle has been circumscribed. Q. E. F.
And it is manifest that, when the centre of the circle falls within the triangle, the angle
BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line
BC, the angle
BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle
BAC, being in a segment less than the semicircle, is greater than a right angle. [
III. 31]
PROPOSITION 6.
In a given circle to inscribe a square.
Let
ABCD be the given circle; thus it is required to inscribe a square in the circle
ABCD.
Let two diameters
AC,
BD of the circle
ABCD be drawn at right angles to one another, and let
AB,
BC,
CD,
DA be joined.
Then, since
BE is equal to
ED, for
E is the centre, and
EA is common and at right angles, therefore the base
AB is equal to the base
AD. [
I. 4]
For the same reason each of the straight lines
BC,
CD is also equal to each of the straight lines
AB,
AD;
therefore the quadrilateral ABCD is equilateral.
I say next that it is also right-angled.
For, since the straight line
BD is a diameter of the circle
ABCD, therefore
BAD is a semicircle;
therefore the angle BAD is right. [III. 31]
For the same reason each of the angles
ABC,
BCD,
CDA is also right;
therefore the quadrilateral ABCD is right-angled.
But it was also proved equilateral; therefore it is a square; [
I. Def. 22] and it has been inscribed in the circle
ABCD.
Therefore in the given circle the square
ABCD has been inscribed. Q. E. F.
PROPOSITION 7.
About a given circle to circumscribe a square.
Let
ABCD be the given circle; thus it is required to circumscribe a square about the circle
ABCD.
Let two diameters
AC,
BD of the circle
ABCD be drawn at right angles to one another, and through the points
A,
B,
C,
D let
FG,
GH,
HK,
KF be drawn touching the circle
ABCD. [
III. 16, Por.]
Then, since
FG touches the circle
ABCD, and
EA has been joined from the centre
E to the point of contact at
A,
therefore the angles at A are right. [III. 18]
For the same reason
the angles at the points B, C, D are also right.
Now, since the angle
AEB is right, and the angle
EBG is also right,
therefore GH is parailel to AC. [I. 28]
For the same reason
AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]
Similarly we can prove that
each of the straight lines GF, HK is parallel to BED.
Therefore
GK,
GC,
AK,
FB,
BK are parallelograms; therefore
GF is equal to
HK, and
GH to
FK. [
I. 34]
And, since
AC is equal to
BD, and
AC is also equal to each of the straight lines
GH,
FK,
while BD is equal to each of the straight lines GF, HK, [I. 34] therefore the quadrilateral FGHK is equilateral.
I say next that it is also right-angled.
For, since
GBEA is a parallelogram, and the angle
AEB is right, therefore the angle
AGB is also right. [
I. 34]
Similarly we can prove that
the angles at H, K, F are also right.
Therefore
FGHK is right-angled.
But it was also proved equilateral;
therefore it is a square; and it has been circumscribed about the circle
ABCD.
Therefore about the given circle a square has been circumscribed. Q. E. F.
PROPOSITION 8.
In a given square to inscribe a circle.
Let
ABCD be the given square; thus it is required to inscribe a circle in the given square
ABCD.
Let the straight lines
AD,
AB be bisected at the points
E,
F respectively [
I. 10], through
E let
EH be drawn parallel to either
AB or
CD, and through
F let
FK be drawn parallel to either
AD or
BC; [
I. 31] therefore each of the figures
AK,
KB,
AH,
HD,
AG,
GC,
BG,
GD is a parallelogram, and their opposite sides are evidently equal. [
I. 34]
Now, since
AD is equal to
AB, and
AE is half of
AD, and
AF half of
AB,
therefore AE is equal to AF, so that the opposite sides are also equal;
therefore FG is equal to GE.
Similarly we can prove that each of the straight lines
GH,
GK is equal to each of the straight lines
FG,
GE;
therefore the four straight lines GE, GF, GH, GK are equal to one another.
Therefore the circle described with centre
G and distance one of the straight lines
GE,
GF,
GH,
GK will pass also through the remaining points.
And it will touch the straight lines
AB,
BC,
CD,
DA, because the angles at
E,
F,
H,
K are right.
For, if the circle cuts
AB,
BC,
CD,
DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [
III. 16] therefore the circle described with centre
G and distance one of the straight lines
GE,
GF,
GH,
GK will not cut the straight lines
AB,
BC,
CD,
DA.
Therefore it will touch them, and will have been inscribed in the square
ABCD.
Therefore in the given square a circle has been inscribed. Q. E. F.
PROPOSITION 9.
About a given square to circumscribe a circle.
Let
ABCD be the given square; thus it is required to circumscribe a circle about the square
ABCD.
For let
AC,
BD be joined, and let them cut one another at
E.
Then, since
DA is equal to
AB, and
AC is common, therefore the two sides
DA,
AC are equal to the two sides
BA,
AC; and the base
DC is equal to the base
BC;
therefore the angle DAC is equal to the angle BAC. [I. 8]
Therefore the angle
DAB is bisected by
AC.
Similarly we can prove that each of the angles
ABC,
BCD,
CDA is bisected by the straight lines
AC,
DB.
Now, since the angle
DAB is equal to the angle
ABC, and the angle
EAB is half the angle
DAB, and the angle
EBA half the angle
ABC, therefore the angle
EAB is also equal to the angle
EBA; so that the side
EA is also equal to
EB. [
I. 6]
Similarly we can prove that each of the straight lines
EA,
EB is equal to each of the straight lines
EC,
ED.
Therefore the four straight lines
EA,
EB,
EC,
ED are equal to one another.
Therefore the circle described with centre
E and distance one of the straight lines
EA,
EB,
EC,
ED will pass also through the remaining points; and it will have been circumscribed about the square
ABCD.
Let it be circumscribed, as
ABCD.
Therefore about the given square a circle has been circumscribed. Q. E. F.
PROPOSITION 10.
To construct an isosceles triangle having each of the angles at the base double of the remaining one.
Let any straight line
AB be set out, and let it be cut at the point
C so that the rectangle contained by
AB,
BC is equal to the square on
CA; [
II. 11] with centre
A and distance
AB let the circle
BDE be described, and let there be fitted in the circle
BDE the straight line
BD equal to the straight line
AC which is not greater than the diameter of the circle
BDE. [
IV. 1]
Let
AD,
DC be joined, and let the circle
ACD be circumscribed about the triangle
ACD. [
IV. 5]
Then, since the rectangle
AB,
BC is equal to the square on
AC, and
AC is equal to
BD, therefore the rectangle
AB,
BC is equal to the square on
BD.
And, since a point
B has been taken outside the circle
ACD, and from
B the two straight lines
BA,
BD have fallen on the circle
ACD, and one of them cuts it, while the other falls on it, and the rectangle
AB,
BC is equal to the square on
BD,
therefore BD touches the circle ACD. [III. 37]
Since, then,
BD touches it, and
DC is drawn across from the point of contact at
D, therefore the angle
BDC is equal to the angle
DAC in the alternate segment of the circle. [
III. 32]
Since, then, the angle
BDC is equal to the angle
DAC, let the angle
CDA be added to each; therefore the whole angle
BDA is equal to the two angles
CDA,
DAC.
But the exterior angle
BCD is equal to the angles
CDA,
DAC; [
I. 32] therefore the angle
BDA is also equal to the angle
BCD.
But the angle
BDA is equal to the angle
CBD, since the side
AD is also equal to
AB; [
I. 5] so that the angle
DBA is also equal to the angle
BCD.
Therefore the three angles
BDA,
DBA,
BCD are equal to one another.
And, since the angle
DBC is equal to the angle
BCD,
the side BD is also equal to the side DC. [I. 6]
But
BD is by hypothesis equal to
CA; therefore
CA is also equal to
CD,
so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles
CDA,
DAC are double of the angle
DAC.
But the angle
BCD is equal to the angles
CDA,
DAC; therefore the angle
BCD is also double of the angle
CAD.
But the angle
BCD is equal to each of the angles
BDA,
DBA; therefore each of the angles
BDA,
DBA is also double of the angle
DAB.
Therefore the isosceles triangle
ABD has been constructed having each of the angles at the base
DB double of the remaining one. Q. E. F.
PROPOSITION 11.
In a given circle to inscribe an equilateral and equiangular pentagon.
Let
ABCDE be the given circle; thus it is required to inscribe in the circle
ABCDE an equilateral and equiangular pentagon.
Let the isosceles triangle
FGH be set out having each of the angles at
G,
H double of the angle at
F; [
IV. 10] let there be inscribed in the circle
ABCDE the triangle
ACD equiangular with the triangle
FGH, so that the angle
CAD is equal to the angle at
F and the angles at
G,
H respectively equal to the angles
ACD,
CDA; [
IV. 2] therefore each of the angles
ACD,
CDA is also double of the angle
CAD.
Now let the angles
ACD,
CDA be bisected respectively by the straight lines
CE,
DB [
I. 9], and let
AB,
BC,
DE,
EA be joined.
Then, since each of the angles
ACD,
CDA is double of the angle
CAD, and they have been bisected by the straight lines
CE,
DB, therefore the five angles
DAC,
ACE,
ECD,
CDB,
BDA are equal to one another.
But equal angles stand on equal circumferences; [
III. 26] therefore the five circumferences
AB,
BC,
CD,
DE,
EA are equal to one another.
But equal circumferences are subtended by equal straight lines; [
III. 29] therefore the five straight lines
AB,
BC,
CD,
DE,
EA are equal to one another;
therefore the pentagon ABCDE is equilateral.
I say next that it is also equiangular.
For, since the circumference
AB is equal to the circumference
DE, let
BCD be added to each;
therefore the whole circumference ABCD is equal to the whole circumference EDCB.
And the angle
AED stands on the circumference
ABCD, and the angle
BAE on the circumference
EDCB;
therefore the angle BAE is also equal to the angle AED. [III. 27]
For the same reason each of the angles
ABC,
BCD,
CDE is also equal to each of the angles
BAE,
AED;
therefore the pentagon ABCDE is equiangular.
But it was also proved equilateral;
therefore in the given circle an equilateral and equiangular pentagon has been inscribed. Q. E. F.
PROPOSITION 12.
About a given circle to circumscribe an equilateral and equiangular pentagon.
Let
ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle
ABCDE.
Let
A,
B,
C,
D,
E be conceived to be the angular points of the inscribed pentagon, so that the circumferences
AB,
BC,
CD,
DE,
EA are equal; [
IV. 11] through
A,
B,
C,
D,
E let
GH,
HK,
KL,
LM,
MG be drawn touching the circle; [
III. 16, Por.] let the centre
F of the circle
ABCDE be taken [
III. 1], and let
FB,
FK,
FC,
FL,
FD be joined.
Then, since the straight line
KL touches the circle
ABCDE at
C, and
FC has been joined from the centre
F to the point of contact at
C,
therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right.
For the same reason
the angles at the points B, D are also right.
And, since the angle
FCK is right, therefore the square on
FK is equal to the squares on
FC,
CK.
For the same reason [
I. 47]
the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on
CK which remains is equal to the square on
BK.
Therefore
BK is equal to
CK.
And, since
FB is equal to
FC, and
FK common,
the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8] and the angle BKF to the angle FKC. Therefore the angle
BFC is double of the angle
KFC,
and the angle BKC of the angle FKC.
For the same reason
the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC.
Now, since the circumference
BC is equal to
CD, the angle
BFC is also equal to the angle
CFD. [
III. 27]
And the angle
BFC is double of the angle
KFC, and the angle
DFC of the angle
LFC;
therefore the angle KFC is also equal to the angle LFC.
But the angle
FCK is also equal to the angle
FCL; therefore
FKC,
FLC are two triangles having two angles equal to two angles and one side equal to one side, namely
FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [
I. 26]
therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC.
And, since
KC is equal to
CL, therefore
KL is double of
KC.
For the same reason it can be proved that
HK is also double of BK.
And
BK is equal to
KC;
therefore HK is also equal to KL.
Similarly each of the straight lines
HG,
GM,
ML can also be proved equal to each of the straight lines
HK,
KL;
therefore the pentagon GHKLM is equilateral.
I say next that it is also equiangular.
For, since the angle
FKC is equal to the angle
FLC, and the angle
HKL was proved double of the angle
FKC,
and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM.
Similarly each of the angles
KHG,
HGM,
GML can also be proved equal to each of the angles
HKL,
KLM; therefore the five angles
GHK,
HKL,
KLM,
LMG,
MGH are equal to one another.
Therefore the pentagon
GHKLM is equiangular.
And it was also proved equilateral; and it has been circumscribed about the circle
ABCDE. Q. E. F.
PROPOSITION 13.
In a given pentagon,
which is equilateral and equiangular,
to inscribe a circle.
Let
ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon
ABCDE.
For let the angles
BCD,
CDE be bisected by the straight lines
CF,
DF respectively; and from the point
F, at which the straight lines
CF,
DF meet one another, let the straight lines
FB,
FA,
FE be joined.
Then, since
BC is equal to
CD, and
CF common, the two sides
BC,
CF are equal to the two sides
DC,
CF; and the angle
BCF is equal to the angle
DCF;
therefore the base BF is equal to the base DF, and the triangle
BCF is equal to the triangle
DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
CBF is equal to the angle
CDF.
And, since the angle
CDE is double of the angle
CDF, and the angle
CDE is equal to the angle
ABC, while the angle
CDF is equal to the angle
CBF; therefore the angle
CBA is also double of the angle
CBF;
therefore the angle ABF is equal to the angle FBC; therefore the angle
ABC has been bisected by the straight line
BF.
Similarly it can be proved that the angles
BAE,
AED have also been bisected by the straight lines
FA,
FE respectively.
Now let
FG,
FH,
FK,
FL,
FM be drawn from the point
F perpendicular to the straight lines
AB,
BC,
CD,
DE,
EA.
Then, since the angle
HCF is equal to the angle
KCF, and the right angle
FHC is also equal to the angle
FKC,
FHC,
FKC are two triangles having two angles equal to two angles and one side equal to one side, namely
FC 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 the perpendicular
FH is equal to the perpendicular
FK.
Similarly it can be proved that each of the straight lines
FL,
FM,
FG is also equal to each of the straight lines
FH,
FK; therefore the five straight lines
FG,
FH,
FK,
FL,
FM are equal to one another.
Therefore the circle described with centre
F and distance one of the straight lines
FG,
FH,
FK,
FL,
FM will pass also through the remaining points; and it will touch the straight lines
AB,
BC,
CD,
DE,
EA, because the angles at the points
G,
H,
K,
L,
M are right.
For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [
III. 16]
Therefore the circle described with centre
F and distance one of the straight lines
FG,
FH,
FK,
FL,
FM will not cut the straight lines
AB,
BC,
CD,
DE,
EA;
therefore it will touch them.
Let it be described, as
GHKLM.
Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed. Q. E. F.
PROPOSITION 14.
About a given pentagon,
which is equilateral and equiangular,
to circumscribe a circle.
Let
ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon
ABCDE.
Let the angles
BCD,
CDE be bisected by the straight lines
CF,
DF respectively, and from the point
F, at which the straight lines meet, let the straight lines
FB,
FA,
FE be joined to the points
B,
A,
E.
Then in manner similar to the preceding it can be proved that the angles
CBA,
BAE,
AED have also been bisected by the straight lines
FB,
FA,
FE respectively.
Now, since the angle
BCD is equal to the angle
CDE, and the angle
FCD is half of the angle
BCD, and the angle
CDF half of the angle
CDE, therefore the angle
FCD is also equal to the angle
CDF,
so that the side FC is also equal to the side FD. [I. 6]
Similarly it can be proved that each of the straight lines
FB,
FA,
FE is also equal to each of the straight lines
FC,
FD; therefore the five straight lines
FA,
FB,
FC,
FD,
FE are equal to one another.
Therefore the circle described with centre
F and distance one of the straight lines
FA,
FB,
FC,
FD,
FE will pass also through the remaining points, and will have been circumscribed.
Let it be circumscribed, and let it be
ABCDE.
Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.
PROPOSITION 15.
In a given circle to inscribe an equilateral and equiangular hexagon.
Let
ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle
ABCDEF.
Let the diameter
AD of the circle
ABCDEF be drawn; let the centre
G of the circle be taken, and with centre
D and distance
DG let the circle
EGCH be described; let
EG,
CG be joined and carried through to the points
B,
F, and let
AB,
BC,
CD,
DE,
EF,
FA be joined.
I say that the hexagon
ABCDEF is equilateral and equiangular.
For, since the point
G is the centre of the circle
ABCDEF,
GE is equal to GD.
Again, since the point
D is the centre of the circle
GCH,
DE is equal to DG.
But
GE was proved equal to
GD;
therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles
EGD,
GDE,
DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [
I. 5]
And the three angles of the triangle are equal to two right angles; [
I. 32]
therefore the angle EGD is one-third of two right angles.
Similarly, the angle
DGC can also be proved to be onethird of two right angles.
And, since the straight line
CG standing on
EB makes the adjacent angles
EGC,
CGB equal to two right angles, therefore the remaining angle
CGB is also one-third of two right angles.
Therefore the angles
EGD,
DGC,
CGB are equal to one another; so that the angles vertical to them, the angles
BGA,
AGF,
FGE are equal. [
I. 15]
Therefore the six angles
EGD,
DGC,
CGB,
BGA,
AGF,
FGE are equal to one another.
But equal angles stand on equal circumferences; [
III. 26] therefore the six circumferences
AB,
BC,
CD,
DE,
EF,
FA are equal to one another.
And equal circumferences are subtended by equal straight lines; [
III. 29]
therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral.
I say next that it is also equiangular.
For, since the circumference
FA is equal to the circumference
ED,
let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA;
and the angle
FED stands on the circumference
FABCD, and the angle
AFE on the circumference
EDCBA;
therefore the angle AFE is equal to the angle DEF. [III. 27]
Similarly it can be proved that the remaining angles of the hexagon
ABCDEF are also severally equal to each of the angles
AFE,
FED;
therefore the hexagon ABCDEF is equiangular.
But it was also proved equilateral; and it has been inscribed in the circle
ABCDEF.
Therefore in the given circle an equilateral and equiangular hexagon has been inscribed. Q. E. F.
PORISM.
From this it is manifest that the side of the hexagon is equal to the radius of the circle.
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.
And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. Q. E. F.
PROPOSITION 16.
In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.
Let
ABCD be the given circle; thus it is required to inscribe in the circle
ABCD a fifteenangled figure which shall be both equilateral and equiangular.
In the circle
ABCD let there be inscribed a side
AC of the equilateral triangle inscribed in it, and a side
AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle
ABCD, there will be five in the circumference
ABC which is one-third of the circle, and there will be three in the circumference
AB which is one-fifth of the circle;
therefore in the remainder BC there will be two of the equal segments.
Let
BC be bisected at
E; [
III. 30] therefore each of the circumferences
BE,
EC is a fifteenth of the circle
ABCD.
If therefore we join
BE,
EC and fit into the circle
ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it. Q. E. F.
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.
And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it. Q. E. F.