PROPOSITION 8.
If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order,
they cut one another in extreme and mean ratio,
and their greater segments are equal to the side of the pentagon.
For in the equilateral and equiangular pentagon
ABCDE let the straight lines
AC,
BE, cutting one another at the point
H, subtend two angles taken in order, the angles at
A,
B; I say that each of them has been cut in extreme and mean ratio at the point
H, and their greater segments are equal to the side of the pentagon.
For let the circle
ABCDE be circumscribed about the pentagon
ABCDE. [
IV. 14]
Then, since the two straight lines
EA,
AB are equal to the two
AB,
BC, and they contain equal angles, therefore the base
BE is equal to the base
AC, the triangle
ABE is equal to the triangle
ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
BAC is equal to the angle
ABE; therefore the angle
AHE is double of the angle
BAH. [
I. 32]
But the angle
EAC is also double of the angle
BAC, inasmuch as the circumference
EDC is also double of the circumference
CB; [
III. 28,
VI. 33] therefore the angle
HAE is equal to the angle
AHE; hence the straight line
HE is also equal to
EA, that is, to
AB. [
I. 6]
And, since the straight line
BA is equal to
AE, the angle
ABE is also equal to the angle
AEB. [
I. 5]
But the angle
ABE was proved equal to the angle
BAH; therefore the angle
BEA is also equal to the angle
BAH.
And the angle
ABE is common to the two triangles
ABE and
ABH; therefore the remaining angle
BAE is equal to the remaining angle
AHB; [
I. 32] therefore the triangle
ABE is equiangular with the triangle
ABH; therefore, proportionally, as
EB is to
BA, so is
AB to
BH. [
VI. 4]
But
BA is equal to
EH; therefore, as
BE is to
EH, so is
EH to
HB.
And
BE is greater than
EH; therefore
EH is also greater than
HB. [
V. 14]
Therefore
BE has been cut in extreme and mean ratio at
H, and the greater segment
HE is equal to the side of the pentagon.
Similarly we can prove that
AC has also been cut in extreme and mean ratio at
H, and its greater segment
CH is equal to the side of the pentagon. Q. E. D.
