parallelogram equality
======================

.. index:: proof, parallelograms

.. image:: elem.1.prop.35.b.png
   :align: right
   :width: 300px

.. image:: elem.1.prop.35.png
   :align: right
   :width: 300px

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

===

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`. [I.34]

For the same reason also 

- `EF` is equal to `BC`, so that `AD` is also equal to `EF`; [I.c.n.1]

and `DE` is common; 

- therefore the whole `AE` is equal to the whole `DF`. [I.c.n.2]

But `AB` is also equal to `DC`; [I.34] therefore the two sides `EA`, `AB` are equal to the two sides `FD`, `DC` respectively,

- and the angle `FDC`[^I.35:1] is equal to the angle `EAB`, the exterior to the interior; [I.29] therefore the base `EB` is equal to the base `FC`, and the triangle `EAB` will be equal to the triangle `FDC`. [I.4]

Let `DGE` be subtracted [^I.35:2] from each; therefore the trapezium `ABGD` which remains is equal to the trapezium `EGCF` which remains. [I.c.n.3] 

Let the triangle `GBC` be added to each; therefore the whole parallelogram `ABCD` is equal to the whole parallelogram `EBCF`. [I.c.n.2]

Therefore etc.

- Q. E. D.


## References

[I.4]: /elem.1.4 "Book 1 - Proposition 4"
[I.29]: /elem.1.29 "Book 1 - Proposition 29"
[I.34]: /elem.1.34 "Book 1 - Proposition 34"
[I.c.n.1]: /elem.1.c.n.1 "Book 1 - Common Notion 1"
[I.c.n.2]: /elem.1.c.n.2 "Book 1 - Common Notion 2"
[I.c.n.3]: /elem.1.c.n.3 "Book 1 - Common Notion 3"

## Footnotes

[^I.35:1]: FDC
    The text has <quote>`DFC`.</quote>


[^I.35:2]: DGE
    Euclid speaks of the triangle `DGE` without any explanation that, in the case which he takes (where `AD`, `EF` have no point in common), `BE`, `CD` must meet at a point `G` between the two parallels. He allows this to appear from the figure simply.