construct a parallelogram equal to a given polygon on a given angle
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.. index:: construction, parallelograms

.. image:: elem.1.prop.45.png
   :align: right
   :width: 300px

To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

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Let `ABCD` be the given rectilineal figure [^I.45:1] and `E` the given rectilineal angle; thus it is required to construct, in the given angle `E`, a parallelogram equal to the rectilineal figure `ABCD`. 

Let `DB` be joined, and let the parallelogram `FH` be constructed equal to the triangle `ABD`, in the angle `HKF` which is equal to `E`; [I.42] let the parallelogram `GM` equal to the triangle `DBC` be applied to the straight line `GH`, in the angle `GHM` which is equal to `E`. [I.44]

Then, since the angle `E` is equal to each of the angles `HKF`, `GHM`, 

- the angle `HKF` is also equal to the angle `GHM`. [I.c.n.1]

Let the angle `KHG` be added to each; therefore the angles `FKH`, `KHG` are equal to the angles `KHG`, `GHM`.

But the angles `FKH`, `KHG` are equal to two right angles; [I.29] therefore the angles `KHG`, `GHM` are also equal to two right angles.

Thus, with a straight line `GH`, and at the point `H` on it, two straight lines `KH`, `HM` not lying on the same side make the adjacent angles equal to two right angles; 

- therefore `KH` is in a straight line with `HM`. [I.14]

And, since the straight line `HG` falls upon the parallels `KM`, `FG`, the alternate angles `MHG`, `HGF` are equal to one another. [I.29]

Let the angle `HGL` be added to each; therefore the angles `MHG`, `HGL` are equal to the angles `HGF`, `HGL`. [I.c.n.2]

But the angles `MHG`, `HGL` are equal to two right angles; [I.29] therefore the angles `HGF`, `HGL` are also equal to two right angles. [I.c.n.1] 

- Therefore `FG` is in a straight line with `GL`. [I.14]

And, since `FK` is equal and parallel to `HG`, [I.34] 

- and `HG` to `ML` also,

`KF` is also equal and parallel to `ML`; [I.c.n.1;1.30] and the straight lines `KM`, `FL` join them (at their extremities); therefore `KM`, `FL` are also equal and parallel. [I.33] 

- Therefore `KFLM` is a parallelogram.

And, since the triangle `ABD` is equal to the parallelogram `FH`, 

- and `DBC` to `GM`,

the whole rectilineal figure `ABCD` is equal to the whole parallelogram `KFLM`.

Therefore the parallelogram `KFLM` has been constructed equal to the given rectilineal figure `ABCD`, in the angle `FKM` which is equal to the given angle `E`.

- Q. E. F.

## References

[I.14]: /elem.1.14 "Book 1 - Proposition 14"
[I.29]: /elem.1.29 "Book 1 - Proposition 29"
[I.30]: /elem.1.30 "Book 1 - Proposition 30"
[I.33]: /elem.1.33 "Book 1 - Proposition 33"
[I.34]: /elem.1.34 "Book 1 - Proposition 34"
[I.42]: /elem.1.42 "Book 1 - Proposition 42"
[I.44]: /elem.1.44 "Book 1 - Proposition 44"
[I.c.n.X]: /elem.1.c.n.X "Book 1 - Common Notion X"
[I.c.n.X]: /elem.1.c.n.X "Book 1 - Common Notion X"

## Footnotes

[^I.45:1]: rectilineal figure, in the Greek
    <quote>rectilineal</quote> simply, without <quote>figure,</quote> <foreign lang="greek">εὐθύγραμμον</foreign> being here used as a substantive, like the similarly formed <foreign lang="greek">παραλληλόγραμμον</foreign>.