---
book: 1
number: 35
id: "I.35"
kind: "theorem"
uses: ["[[book-1/proposition-34]]", "[[book-1/proposition-29]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.35"
license: "CC-BY-SA-4.0"
---

# I.35

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

![I.35](figures/I-35.svg)

## Proof

Let *ABCD*, *EBCF* be parallelograms on the same base *BC* and in the same parallels *AF*, *BC*; I say that *ABCD* is equal to the parallelogram *EBCF*.

For, since *ABCD* is a parallelogram, *AD* is equal to *BC*. [[book-1/proposition-34|I. 34]]

For the same reason also *EF* is equal to *BC*, so that *AD* is also equal to *EF*; [[book-1/common-notions#Common Notion 1|C.N. 1]] and *DE* is common; therefore the whole *AE* is equal to the whole *DF*. [[book-1/common-notions#Common Notion 2|C.N. 2]]

But *AB* is also equal to *DC*; [[book-1/proposition-34|I. 34]] therefore the two sides *EA*, *AB* are equal to the two sides *FD*, *DC* respectively, and the angle *FDC* is equal to the angle *EAB*, the exterior to the interior; [[book-1/proposition-29|I. 29]] therefore the base *EB* is equal to the base *FC*, and the triangle *EAB* will be equal to the triangle *FDC*. [[book-1/proposition-4|I. 4]]

Let *DGE* be subtracted from each; therefore the trapezium *ABGD* which remains is equal to the trapezium *EGCF* which remains. [[book-1/common-notions#Common Notion 3|C.N. 3]]

Let the triangle *GBC* be added to each; therefore the whole parallelogram *ABCD* is equal to the whole parallelogram *EBCF*. [[book-1/common-notions#Common Notion 2|C.N. 2]]

Therefore etc.

Q. E. D.