** Proposition 7**. *If three angles of an equilateral
pentagon, taken either in order or not in order, be equal, the pentagon
will be equiangular.*

For in the equilateral pentagon ABCDE let, first, three angles taken
in order, those at A, B, C, be equal to one another; I say that
the pentagon ABCDE is equiangular.

For let AC, BE, FD be joined.

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] [p. 452]

But the whole AC was also proved equal to the whole BE; therefore
the remainder FC is also equal to the remainder FE.

But CD is also equal to DE.

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]

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.

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.

Similarly we can prove that the angle CDE is also equal to the angles
at A, B, C; therefore the pentagon ABCDE is equiangular.

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.

For let BD be joined.

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.

But the angle BED is also equal to the angle BDE, since the side
BE is also equal to the side BD. [I. 5]

Therefore the whole angle AED is equal to the whole angle CDE.

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. [p.
453]

For the same reason the angle ABC is also equal to the angles at
A, C, D.

Therefore the pentagon ABCDE is equiangular.

Q. E. D.