---
book: 1
number: 39
id: "I.39"
kind: "theorem"
uses: ["[[book-1/proposition-31]]", "[[book-1/proposition-37]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.39"
license: "CC-BY-SA-4.0"
---

# I.39

Equal triangles which are on the same base and on the same side are also in the same parallels.

![I.39](figures/I-39.svg)

## Proof

Let *ABC*, *DBC* be equal triangles which are on the same base *BC* and on the same side of it; [I say that they are also in the same parallels.]

And [For] let *AD* be joined; I say that *AD* is parallel to *BC*.

For, if not, let *AE* be drawn through the point *A* parallel to the straight line *BC*, [[book-1/proposition-31|I. 31]] and let *EC* be joined.

Therefore the triangle *ABC* is equal to the triangle *EBC*; for it is on the same base *BC* with it and in the same parallels. [[book-1/proposition-37|I. 37]]

But *ABC* is equal to *DBC*; therefore *DBC* is also equal to *EBC*, [[book-1/common-notions#Common Notion 1|C.N. 1]] the greater to the less: which is impossible.

Therefore *AE* is not parallel to *BC*.

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

Therefore etc.

Q. E. D.