---
book: 1
number: 36
id: "I.36"
kind: "theorem"
uses: ["[[book-1/proposition-33]]", "[[book-1/proposition-34]]", "[[book-1/proposition-35]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.36"
license: "CC-BY-SA-4.0"
---

# I.36

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

![I.36](figures/I-36.svg)

## Proof

Let *ABCD*, *EFGH* be parallelograms which are on equal bases *BC*, *FG* and in the same parallels *AH*, *BG*; I say that the parallelogram *ABCD* is equal to *EFGH*.

For let *BE*, *CH* be joined.

Then, since *BC* is equal to *FG* while *FG* is equal to *EH*, *BC* is also equal to *EH*. [[book-1/common-notions#Common Notion 1|C.N. 1]]

But they are also parallel.

And *EB*, *HC* join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [[book-1/proposition-33|I. 33]]

Therefore *EBCH* is a parallelogram. [[book-1/proposition-34|I. 34]]

And it is equal to *ABCD*; for it has the same base *BC* with it, and is in the same parallels *BC*, *AH* with it. [[book-1/proposition-35|I. 35]]

For the same reason also *EFGH* is equal to the same *EBCH*; [[book-1/proposition-35|I. 35]] so that the parallelogram *ABCD* is also equal to *EFGH*. [[book-1/common-notions#Common Notion 1|C.N. 1]]

Therefore etc. Q. E. D.