Book 13
7
Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB.
7
But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5]
7
Therefore the whole angle AED is equal to the whole angle CDE.
7
But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.
7
For the same reason the angle ABC is also equal to the angles at A, C, D.
7
Therefore the pentagon ABCDE is equiangular. Q. E. D.
PROPOSITION 8.
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.
8
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.
8
For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]
