PROPOSITION 10.
If an equilateral pentagon be inscribed in a circle,
the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.
Let
ABCDE be a circle, and let the equilateral pentagon
ABCDE be inscribed in the circle
ABCDE.
I say that the square on the side of the pentagon
ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle
ABCDE.
For let the centre of the circle, the point
F, be taken, let
AF be joined and carried through to the point
G, let
FB be joined, let
FH be drawn from
F perpendicular to
AB and be carried through to
K, let
AK,
KB be joined, let
FL be again drawn from
F perpendicular to
AK, and be carried through to
M, and let
KN be joined.
Since the circumference
ABCG is equal to the circumference
AEDG, and in them
ABC is equal to
AED, therefore the remainder, the circumference
CG, is equal to the remainder
GD.
But
CD belongs to a pentagon; therefore
CG belongs to a decagon.
And, since
FA is equal to
FB, and
FH is perpendicular, therefore the angle
AFK is also equal to the angle
KFB. [
I. 5,
I. 26]
Hence the circumference
AK is also equal to
KB; [
III. 26] therefore the circumference
AB is double of the circumference
BK; therefore the straight line
AK is a side of a decagon.
For the same reason
AK is also double of
KM.
Now, since the circumference
AB is double of the circumference
BK, while the circumference
CD is equal to the circumference
AB, therefore the circumference
CD is also double of the circumference
BK.
But the circumference
CD is also double of
CG; therefore the circumference
CG is equal to the circumference
BK.
But
BK is double of
KM, since
KA is so also; therefore
CG is also double of
KM.
But, further, the circumference
CB is also double of the circumference
BK, for the circumference
CB is equal to
BA.
Therefore the whole circumference
GB is also double of
BM; hence the angle
GFB is also double of the angle
BFM. [
VI. 33]
But the angle
GFB is also double of the angle
FAB, for the angle
FAB is equal to the angle
ABF.
Therefore the angle
BFN is also equal to the angle
FAB.
But the angle
ABF is common to the two triangles
ABF and
BFN; therefore the remaining angle
AFB is equal to the remaining angle
BNF; [
I. 32] therefore the triangle
ABF is equiangular with the triangle
BFN.
Therefore, proportionally, as the straight line
AB is to
BF, so is
FB to
BN; [
VI. 4] therefore the rectangle
AB,
BN is equal to the square on
BF. [
VI. 17]
Again, since
AL is equal to
LK, while
LN is common and at right angles, therefore the base
KN is equal to the base
AN; [
I. 4] therefore the angle
LKN is also equal to the angle
LAN.
But the angle
LAN is equal to the angle
KBN; therefore the angle
LKN is also equal to the angle
KBN.
And the angle at
A is common to the two triangles
AKB and
AKN.
Therefore the remaining angle
AKB is equal to the remaining angle
KNA; [
I. 32] therefore the triangle
KBA is equiangular with the triangle
KNA.
Therefore, proportionally, as the straight line
BA is to
AK, so is
KA to
AN; [
VI. 4] therefore the rectangle
BA,
AN is equal to the square on
AK. [
VI. 17]
But the rectangle
AB,
BN was also proved equal to the square on
BF; therefore the rectangle
AB,
BN together with the rectangle
BA,
AN, that is, the square on
BA [
II. 2], is equal to the square on
BF together with the square on
AK.
And
BA is a side of the pentagon,
BF of the hexagon [
IV. 15, Por.], and
AK of the decagon.
Therefore etc. Q. E. D.