triangles and parallels

../../../../_images/elem.1.prop.39.png

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.

## 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.