triangles and parallels
=======================

.. index:: proof, triangles, parallels

.. image:: elem.1.prop.39.png
   :align: right
   :width: 300px

Equal triangles which are on the same base and on the same side are also in the same parallels.

===

Let `ABC`, `DBC` be equal triangles which are on the same base `BC` and on the same side of it; [I say that they are also in the same parallels.] [^I.39:1]

And [For] let `AD` be joined; I say that `AD` is parallel to `BC`.

For, if not, let `AE` be drawn through the point `A` parallel to the straight line `BC`, [I.31] and let `EC` be joined. 

Therefore the triangle `ABC` is equal to the triangle `EBC`; for it is on the same base `BC` with it and in the same parallels. [I.37]

But `ABC` is equal to `DBC`; 

- therefore `DBC` is also equal to `EBC`, [I.c.n.1] the greater to the less: which is impossible.

Therefore `AE` is not parallel to `BC`. 

Similarly we can prove that neither is any other straight line except `AD`; 

- therefore `AD` is parallel to `BC`.

Therefore etc.

- Q. E. D.

## References

[I.31]: /elem.1.31 "Book 1 - Proposition 31"
[I.37]: /elem.1.37 "Book 1 - Proposition 37"
[I.c.n.1]: /elem.1.c.n.1 "Book 1 - Common Notion 1"

## Footnotes

[^I.39:1]: [I say that they are also in the same parallels.]
    Heiberg has proved (<title>Hermes</title>, XXXVIII., 1903, p. 50) from a recently discovered papyrus-fragment (<title>Fayūm towns and their papyri</title>, p. 96, No. IX.) that these words are an interpolation by some one who did not observe that the words <quote>And let `AD` be joined</quote> are part of the <em>setting-out</em> (<foreign lang="greek">ἔκθεσις</foreign>), but took them as belonging to the <em>construction</em> (<foreign lang="greek">κατασκευή</foreign>) and consequently thought that a <foreign lang="greek">διορισμός</foreign> or <quote>definition</quote> (of the thing to be proved) should precede. The interpolator then altered <quote>And</quote> into <quote>For</quote> in the next sentence.