---
book: 1
number: 40
id: "I.40"
kind: "theorem"
uses: ["[[book-1/proposition-31]]", "[[book-1/proposition-38]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.40"
license: "CC-BY-SA-4.0"
---

# I.40

*Equal triangles which are on equal bases and on the same side are also in the same parallels*.

![I.40](figures/I-40.svg)

## Proof

Let *ABC*, *CDE* be equal triangles on equal bases *BC*, *CE* and on the same side.

I say that they are also in the same parallels.

For let *AD* be joined; I say that *AD* is parallel to *BE*.

For, if not, let *AF* be drawn through *A* parallel to *BE* [[book-1/proposition-31|I. 31]], and let *FE* be joined.

Therefore the triangle *ABC* is equal to the triangle *FCE*; for they are on equal bases *BC*, *CE* and in the same parallels *BE*, *AF*. [[book-1/proposition-38|I. 38]]

But the triangle *ABC* is equal to the triangle *DCE*; therefore the triangle *DCE* is also equal to the triangle *FCE*, [[book-1/common-notions#Common Notion 1|C.N. 1]] the greater to the less: which is impossible. Therefore *AF* is not parallel to *BE*.

Similarly we can prove that neither is any other straight line except *AD*; therefore *AD* is parallel to *BE*.

Therefore etc. Q. E. D.]