---
book: 1
number: 37
id: "I.37"
kind: "theorem"
uses: ["[[book-1/proposition-31]]", "[[book-1/proposition-35]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.37"
license: "CC-BY-SA-4.0"
---

# I.37

Triangles which are on the same base and in the same parallels are equal to one another.

![I.37](figures/I-37.svg)

## Proof

Let *ABC*, *DBC* be triangles on the same base *BC* and in the same parallels *AD*, *BC*; I say that the triangle *ABC* is equal to the triangle *DBC*.

Let *AD* be produced in both directions to *E*, *F*; through *B* let *BE* be drawn parallel to *CA*, [[book-1/proposition-31|I. 31]] and through *C* let *CF* be drawn parallel to *BD*. [[book-1/proposition-31|I. 31]]

Then each of the figures *EBCA*, *DBCF* is a parallelogram; and they are equal,

for they are on the same base *BC* and in the same parallels *BC*, *EF*. [[book-1/proposition-35|I. 35]]

Moreover the triangle *ABC* is half of the parallelogram *EBCA*; for the diameter *AB* bisects it. [[book-1/proposition-34|I. 34]]

And the triangle *DBC* is half of the parallelogram *DBCF*; for the diameter *DC* bisects it. [[book-1/proposition-34|I. 34]]

[But the halves of equal things are equal to one another.]

Therefore the triangle *ABC* is equal to the triangle *DBC*.

Therefore etc.

Q. E. D.