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 < A B C D E > let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon < A B C D E > is equiangular.
For let / A C /, / B E /, / F D / be joined.
Now, since the two sides / C B /, / B A / are equal to the two sides / B A /, / A E / respectively, and the angle < C B A ) is equal to the angle < B A E ),
therefore the base / A C / is equal to the base / B E /, the triangle< A B C > is equal to the triangle < A B E >, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [G.24] that is, the angle < B C A ) to the angle < B E A ), and the angle < A B E ) to the angle < C A B );
[G.27]hence the side / A F / is also equal to the side / B F /.
But the whole / A C / was also proved equal to the whole / B E /;
therefore the remainder / F C / is also equal to the remainder / F E /.
But / C D / is also equal to / D E /. Therefore the two sides / F C /, / C D / are equal to the two sides / F E /, / E D /; and the base / F D / is common to them;
[G.29]therefore the angle < F C D ) is equal to the angle < F E D ).
But the angle < B C A ) was also proved equal to the angle < A E B );
therefore the whole angle < B C D ) is also equal to the whole angle < A E D ).
But, by hypothesis, the angle < B C D ) is equal to the angles at A, B;
therefore the angle < A E D ) is also equal to the angles at A, B.
Similarly we can prove that the angle < C D E ) is also equal to the angles at A, B, C; therefore the pentagon < A B C D E > is equiangular.
Next, let the given equal angles not be angles taken in order, but let the angles at the pointsA, C, D be equal; I say that in this case too the pentagon < A B C D E > is equiangular.
For let / B D / be joined.
Then, since the two sides / B A /, / A E / are equal to the two sides / B C /, / C D /, and they contain equal angles,
therefore the base / B E / is equal to the base / B D /, the triangle < A B E > is equal to the triangle < B C D >, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [G.24] therefore the angle < A E B ) is equal to the angle < C D B ).
But the angle < B E D ) is also equal to the angle < B D E ), since the side / B E / is also equal to the side / B D /. [G.25]
Therefore the whole angle < A E D ) is equal to the whole angle < C D E ).
But the angle < C D E ) is, by hypothesis, equal to the angles at A, C;
therefore the angle < A E D ) is also equal to the angles at A, C.
For the same reason the angle < A B C ) is also equal to the angles at A, C, D. Therefore the pentagon < A B C D E > is equiangular.
