---
book: 1
number: 6
id: "I.6"
kind: "theorem"
uses: []
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.6"
license: "CC-BY-SA-4.0"
---

# I.6

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

![I.6](figures/I-6.svg)

## Proof

Let *ABC* be a triangle having the angle *ABC* equal to the angle *ACB*;

I say that the side *AB* is also equal to the side *AC*.

For, if *AB* is unequal to *AC*, one of them is greater.

Let *AB* be greater; and from *AB* the greater let *DB* be cut off equal to *AC* the less;

let *DC* be joined.

Then, since *DB* is equal to *AC*, and *BC* is common, the two sides *DB*, *BC* are equal to the two sides *AC*, *CB* respectively; and the angle *DBC* is equal to the angle *ACB*; therefore the base *DC* is equal to the base *AB*, and the triangle *DBC* will be equal to the triangle *ACB*, the less to the greater: which is absurd. Therefore *AB* is not unequal to *AC*; it is therefore equal to it.

Therefore etc.

Q. E. D.