An Assumpt.
But now that the angle of an equilater and equiangle pentagon is a right angle and a fi••th par•• more of a right angle, may thus be proued. Suppose that ABCDE be an equilater and equiangle pentagon. And (by the 14. of the fourth) describe about it a circle ABCDE. And take (by the 1. of the third) the center thereof,
and let the same be F. And draw these right lines FA, FB, FC, FD, FE. Wherefore those lines do diuide the angles of the pentagon into two equall partes in the poyntes A, B, C, D, E, by the 4. of the first. And
••orasmuch as the fiue angles that are at the poynt F a
••e equall to fower right angles, by the corollary of the 15. of the first, and they are equall the one to the other by the 8. of the first: therfore one of those angles, as
••or example sake, the angle AFB is a fi
••th part lesse then a right angle. Wherfore the angles remayning, namely, FAB, & ABF, are one right angle and a fifth part ouer. But the angle F∣AB is equall to the angle FBC. Wherefore the whole angle ABC being one of the an∣gles of the pentagon is a right angle and a fifth part more then a right angle: which was re∣quired to be proued.