---
book: 13
number: 7
id: "XIII.7"
kind: "theorem"
uses: ["[[book-1/proposition-4]]", "[[book-1/proposition-6]]", "[[book-1/proposition-8]]", "[[book-1/proposition-5]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.7"
license: "CC-BY-SA-4.0"
---

# XIII.7

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

## Proof

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, [[book-1/proposition-4|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*. [[book-1/proposition-6|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*. [[book-1/proposition-8|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; [[book-1/proposition-4|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*. [[book-1/proposition-5|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.