PROPOSITION 9.
If the side of the hexagon and that of the decagon inscribed in the same circle be added together,
the whole straight line has been cut in extreme and mean ratio,
and its greater segment is the side of the hexagon.
Let
ABC be a circle; of the figures inscribed in the circle
ABC let
BC be the side of a decagon,
CD that of a hexagon, and let them be in a straight line; I say that the whole straight line
BD has been cut in extreme and mean ratio, and
CD is its greater segment.
For let the centre of the circle, the point
E, be taken, let
EB,
EC,
ED be joined, and let
BE be carried through to
A.
Since
BC is the side of an equilateral decagon, therefore the circumference
ACB is five times the circumference
BC; therefore the circumference
AC is quadruple of
CB.
But, as the circumference
AC is to
CB, so is the angle
AEC to the angle
CEB; [
VI. 33] therefore the angle
AEC is quadruple of the angle
CEB.
And, since the angle
EBC is equal to the angle
ECB, [
I. 5] therefore the angle
AEC is double of the angle
ECB. [
I. 32]
And, since the straight line
EC is equal to
CD, for each of them is equal to the side of the hexagon inscribed in the circle
ABC, [
IV. 15, Por.] the angle
CED is also equal to the angle
CDE; [
I. 5] therefore the angle
ECB is double of the angle
EDC. [
I. 32]
But the angle
AEC was proved double of the angle
ECB; therefore the angle
AEC is quadruple of the angle
EDC.
But the angle
AEC was also proved quadruple of the angle
BEC; therefore the angle
EDC is equal to the angle
BEC.
But the angle
EBD is common to the two triangles
BEC and
BED; therefore the remaining angle
BED is also equal to the remaining angle
ECB; [
I. 32] therefore the triangle
EBD is equiangular with the triangle
EBC.
Therefore, proportionally, as
DB is to
BE, so is
EB to
BC. [
VI. 4]
But
EB is equal to
CD.
Therefore, as
BD is to
DC, so is
DC to
CB.
And
BD is greater than
DC; therefore
DC is also greater than
CB.
Therefore the straight line
BD has been cut in extreme and mean ratio, and
DC is its greater segment. Q. E. D.