About a given pentagon, which is equilateral and equiangular, to circumscribe a circle.
Let ABCDE be the given pentagon, which is equilateral
and equiangular;
thus it is required to circumscribe a circle
about the pentagon ABCDE.
Let the anglesBCD, CDE be bisected
by the straight linesCF, DF respectively,
and from the pointF, at which the straight
lines meet, let the straight lines FB, FA,
FE be joined to the points B, A, E.
Then in manner similar to the preceding
it can be proved that the angles
CBA, BAE, AED have also been bisected by the straight
lines FB, FA, FE respectively.
Now, since the angle BCD is equal to the angle CDE,
and the angle FCD is half of the angle BCD,
and the angle CDF half of the angle CDE,
therefore the angle FCD is also equal to the angle CDF,
[I. 6]so that the side FC is also equal to the side FD.
Similarly it can be proved that
each of the straight lines FB, FA, FE is also equal to each
of the straight lines FC, FD;
therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.
Therefore the circle described with centreF and distance
one of the straight lines FA, FB, FC, FD, FE will pass
also through the remaining points, and will have been
circumscribed.
Let it be circumscribed, and let it be ABCDE.
Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.
