equality of parallelogram complements
=====================================

.. index:: proof, parallelograms

.. image:: elem.1.prop.43.png
   :align: right
   :width: 300px

In any parallelogram the complements [^I.43:1]  of the parallelograms about the diameter are equal to one another.

===

Let `ABCD` be a parallelogram, and `AC` its diameter; and about `AC` let `EH`, `FG` be parallelograms, and `BK`, `KD` [^I.43:2] the so-called complements;

I say that the complement `BK` is equal to the complement `KD`.

For, since `ABCD` is a parallelogram, and `AC` its diameter, 

- the triangle `ABC` is equal to the triangle `ACD`. [I.34]

Again, since `EH` is a parallelogram, and `AK` is its diameter, 

- the triangle `AEK` is equal to the triangle `AHK`.

For the same reason 

- the triangle `KFC` is also equal to `KGC`.

Now, since the triangle `AEK` is equal to the triangle `AHK`, 

- and `KFC` to `KGC`,

the triangle `AEK` together with `KGC` is equal to the triangle `AHK` together with `KFC`. [I.c.n.2]

And the whole triangle `ABC` is also equal to the whole `ADC`; therefore the complement `BK` which remains is equal to the complement `KD` which remains. [I.c.n.3]

Therefore etc.

- Q. E. D.

## References

[I.34]: /elem.1.34 "Book 1 - Proposition 34"
[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.43:1]: complements,
    <foreign lang="greek">παραπληρώματα</foreign>, the figures put in to fill up (interstices).

[^I.43:2]: and about AC....
    Euclid's phraseology here and in the next proposition implies that the complements as well as the other parallelograms are <quote>about</quote> the diagonal. The words are here <foreign lang="greek">περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΕΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ</foreign>. The expression <quote>the so-called complements</quote> indicates that this technical use of <foreign lang="greek">παραπληρώματα</foreign> was not new, though it might not be universally known.