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

.. image:: elem.1.prop.42.png
   :align: right
   :width: 300px

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

===

Let `ABC` be the given triangle, and `D` the given rectilineal angle; thus it is required to construct in the rectilineal angle `D` a parallelogram equal to the triangle `ABC`. 

Let `BC` be bisected at `E`, and let `AE` be joined; on the straight line `EC`, and at the point `E` on it, let the angle `CEF` be constructed equal to the angle `D`; [I.23] through `A` let `AG` be drawn parallel to `EC`, and [I.31] through `C` let `CG` be drawn parallel to `EF`.

Then `FECG` is a parallelogram.

And, since `BE` is equal to `EC`, 

- the triangle `ABE` is also equal to the triangle `AEC`, for they are on equal bases `BE`, `EC` and in the same parallels `BC`, `AG`; [I.38] therefore the triangle `ABC` is double of the triangle `AEC`.

But the parallelogram `FECG` is also double of the triangle `AEC`, for it has the same base with it and is in the same parallels with it; [I.41] 

- therefore the parallelogram `FECG` is equal to the triangle `ABC`.

And it has the angle `CEF` equal to the given angle `D`.

Therefore the parallelogram `FECG` has been constructed equal to the given triangle `ABC`, in the angle `CEF` which is equal to `D`. 

- Q. E. F.

## References

[I.23]: /elem.1.23 "Book 1 - Proposition 23"
[I.31]: /elem.1.31 "Book 1 - Proposition 31"
[I.38]: /elem.1.38 "Book 1 - Proposition 38"
[I.41]: /elem.1.41 "Book 1 - Proposition 41"