Book 13
7
Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6]
7
But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE.
7
But CD is also equal to DE.
7
Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8]
7
But the angle BCA was also proved equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED.
7
But, by hypothesis, the angle BCD is equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B.
7
Similarly we can prove that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular.
7
Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular.
7
For let BD be joined.
