---
book: 1
number: 38
id: "I.38"
kind: "theorem"
uses: ["[[book-1/proposition-31]]", "[[book-1/proposition-36]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.38"
license: "CC-BY-SA-4.0"
---

# I.38

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

![I.38](figures/I-38.svg)

## Proof

Let *ABC*, *DEF* be triangles on equal bases *BC*, *EF* and in the same parallels *BF*, *AD*; I say that the triangle *ABC* is equal to the triangle *DEF*.

For let *AD* be produced in both directions to *G*, *H*; through *B* let *BG* be drawn parallel to *CA*, [[book-1/proposition-31|I. 31]] and through *F* let *FH* be drawn parallel to *DE*.

Then each of the figures *GBCA*, *DEFH* is a parallelogram; and *GBCA* is equal to *DEFH*;

for they are on equal bases *BC*, *EF* and in the same parallels *BF*, *GH*. [[book-1/proposition-36|I. 36]]

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

And the triangle *FED* is half of the parallelogram *DEFH*; for the diameter *DF* 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 *DEF*.

Therefore etc.

Q. E. D.