---
book: 1
number: 16
id: "I.16"
kind: "theorem"
uses: ["[[book-1/proposition-10]]", "[[book-1/proposition-3]]", "[[book-1/proposition-15]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.16"
license: "CC-BY-SA-4.0"
---

# I.16

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

![I.16](figures/I-16.svg)

## Proof

Let *ABC* be a triangle, and let one side of it *BC* be produced to *D*;

I say that the exterior angle *ACD* is greater than either of the interior and opposite angles *CBA*, *BAC*.

Let *AC* be bisected at *E* [[book-1/proposition-10|I. 10]], and let *BE* be joined and produced in a straight line to *F*;

let *EF* be made equal to *BE*[[book-1/proposition-3|I. 3]], let *FC* be joined [[book-1/postulates#Postulate 1|Post. 1]], and let *AC* be drawn through to *G* [[book-1/postulates#Postulate 2|Post. 2]].

Then, since *AE* is equal to *EC*, and *BE* to *EF*, the two sides *AE*, *EB* are equal to the two sides *CE*, *EF* respectively; and the angle *AEB* is equal to the angle *FEC*, for they are vertical angles. [[book-1/proposition-15|I. 15]] Therefore the base *AB* is equal to the base *FC*, and the triangle *ABE* is equal to the triangle *CFE*, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [[book-1/proposition-4|I. 4]] therefore the angle *BAE* is equal to the angle *ECF*.

But the angle *ECD* is greater than the angle *ECF*; [*C. N*. 5] therefore the angle *ACD* is greater than the angle *BAE*.

Similarly also, if *BC* be bisected, the angle *BCG*, that is, the angle *ACD* [[book-1/proposition-15|I. 15]], can be proved greater than the angle *ABC* as well.

Therefore etc.

Q. E. D.