---
book: 1
number: 18
id: "I.18"
kind: "theorem"
uses: ["[[book-1/proposition-3]]", "[[book-1/proposition-16]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.18"
license: "CC-BY-SA-4.0"
---

# I.18

In any triangle the greater side subtends the greater angle.

![I.18](figures/I-18.svg)

## Proof

For let *ABC* be a triangle having the side *AC* greater than *AB*;

I say that the angle *ABC* is also greater than the angle *BCA*.

For, since *AC* is greater than *AB*, let *AD* be made equal to *AB* [[book-1/proposition-3|I. 3]], and let *BD* bejoined.

Then, since the angle *ADB* is an exterior angle of the triangle *BCD*,

it is greater than the interior and opposite angle *DCB*. [[book-1/proposition-16|I. 16]]

But the angle *ADB* is equal to the angle *ABD*, since the side *AB* is equal to *AD*; therefore the angle *ABD* is also greater than the angle *ACB*; therefore the angle *ABC* is much greater than the angle *ACB*.

Therefore etc.

Q. E. D.