---
book: 1
number: 20
id: "I.20"
kind: "theorem"
uses: ["[[book-1/proposition-5]]", "[[book-1/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.20"
license: "CC-BY-SA-4.0"
---

# I.20

In any triangle two sides taken together in any manner are greater than the remaining one.

![I.20](figures/I-20.svg)

## Proof

For let *ABC* be a triangle; I say that in the triangle *ABC* two sides taken together in any manner are greater than the remaining one, namely *BA*, *AC* greater than *BC*, *AB*, *BC* greater than *AC*, *BC*, *CA* greater than *AB*.

For let *BA* be drawn through to the point *D*, let *DA* be made equal to *CA*, and let *DC* be joined.

Then, since *DA* is equal to *AC*, the angle *ADC* is also equal to the angle *ACD*; [[book-1/proposition-5|I. 5]] therefore the angle *BCD* is greater than the angle *ADC*. [[book-1/common-notions#Common Notion 5|C.N. 5]]

And, since *DCB* is a triangle having the angle *BCD* greater than the angle *BDC*, and the greater angle is subtended by the greater side, [[book-1/proposition-19|I. 19]] therefore *DB* is greater than *BC*.

But *DA* is equal to *AC*; therefore *BA*, *AC* are greater than *BC*.

Similarly we can prove that *AB*, *BC* are also greater than *CA*, and *BC*, *CA* than *AB*.

Therefore etc.

Q. E. D.