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]
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.
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.