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.
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.
gh
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.
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.
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.
Since then ABE, FGL are two triangles having one angle
equal to one angle and the sides about the equal angles proportional, therefore
the triangle ABE is equiangular with the triangle FGL; [VI.
6] so that it is also similar; [VI. 4 and Def.
1] therefore the angle ABE is equal to the angle FGL.
But the whole angle ABC is also equal to the whole angle FGH
because of the similarity of the polygons; therefore the remaining angle
EBC
is equal to the angle LGH.
And, since, because of the similarity of the triangles ABE,
FGL,
as EB is to BA, so is LG to GF, and moreover
also, because of the similarity of the polygons, as AB is
to BC, so is FG to GH, therefore, ex aequali, as EB
is to BC, so is LG to
GH; [V. 22]
that is, the sides about the equal angles EBC,
LGH are proportional;
therefore the triangle
EBC is equiangular with the triangle LGH,
[VI. 6] so that the triangle EBC is also
similar to the triangle
LGH. [VI. 4 and Def. 1]
For the same reason the triangle ECD is also similar to the triangle
LHK.
Therefore the similar polygons ABCDE, FGHKL have been
divided into similar triangles, and into triangles equal in multitude.
I say that they are also in the same ratio as the wholes, that is,
in such manner that the triangles are proportional, and ABE, EBC,
ECD
are antecedents, while FGL, LGH, LHK are their consequents,
and that the polygon ABCDE has to the polygon
FGHKL a ratio
duplicate of that which the corresponding side has to the corresponding
side, that is AB to FG.
For let AC, FH be joined.
Then since, because of the similarity of the polygons, the angle ABC
is equal to the angle FGH, and, as AB is to BC, so
is FG to GH, the triangle ABC is equiangular
with the triangle FGH; [VI. 6] therefore the
angle BAC is equal to the angle GFH, and the angle BCA
to the angle GHF.
And, since the angle BAM is equal to the angle GFN, and
the angle ABM is also equal to the angle FGN, therefore the
remaining angle AMB is also equal to the remaining angle FNG;
[I. 32] therefore the triangle ABM is equiangular
with the triangle FGN.
Similarly we can prove that the triangle BMC is also equiangular
with the triangle GNH.
Therefore, proportionally, as AM is to MB, so is FN
to NG, and, as BM is to MC, so is GN to NH;
so that, in addition, ex aequali, as AM is to MC, so
is FN to NH.
But, as AM is to MC, so is the triangle ABM to
MBC,
and AME to EMC; for they are to one another as their bases.
[VI. 1]
Therefore also, as one of the antecedents is to one of the consequents,
so are all the antecedents to all the consequents; [V. 12]
therefore, as the triangle AMB is to BMC, so is ABE
to CBE.
But, as AMB is to BMC, so is AM to MC;
therefore also, as AM is to MC, so is the triangle ABE
to the triangle EBC.
For the same reason also, as FN is to NH, so is the triangle
FGL
to the triangle GLH.
And, as AM is to MC, so is FN to NH; therefore
also, as the triangle ABE is to the triangle BEC, so is the
triangle FGL to the triangle GLH; and, alternately, as the
triangle ABE is to the triangle FGL, so is the triangle BEC
to the triangle GLH.
Similarly we can prove, if BD, GK be joined, that, as
the triangle BEC is to the triangle LGH, so also is the triangle
ECD
to the triangle LHK.
And since, as the triangle ABE is to the triangle FGL,
so is EBC to LGH, and further ECD to LHK, therefore
also, as one of the antecedents is to one of the consequents so are all
the antecedents to all the consequents; [V. 12 therefore,
as the triangle ABE is to the triangle FGL, so is the polygon
ABCDE
to the polygon FGHKL.
But the triangle ABE has to the triangle FGL a ratio
duplicate of that which the corresponding side AB has to the corresponding
side FG; for similar triangles are in the duplicate ratio of the
corresponding sides. [VI. 19]
Therefore the polygon ABCDE also has to the polygon FGHKL
a ratio duplicate of that which the corresponding side AB has to
the corresponding side FG.
Therefore etc.
But GH is similar to EL; therefore MN is
also similar to EL; [VI. 21] therefore EL
is about the same diameter with MN. [VI. 26]
Let their diameter FO be drawn, and let the figure be described.
Since GH is equal to EL, C, while GH is
equal to MN, therefore MN is also equal to EL, C.
Let EL be subtracted from each; therefore the remainder,
the gnomon XWV, is equal to C.
Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43].
Let EO be added to each; therefore the whole AO is equal to the gnomon VWX.
But the gnomon VWX is equal to C; therefore AO is also equal to C.
Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]. Q. E. F.
