parallelogram equality

../../../../_images/elem.1.prop.35.b.png ../../../../_images/elem.1.prop.35.png

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

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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.

## 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.