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.
