triangles in parallels equality

Triangles which are on the same base and in the same parallels are equal to one another.

===

Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC; I say that the triangle ABC is equal to the triangle DBC.

Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I.31] and through C let CF be drawn parallel to BD. [I.31]

Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,

for they are on the same base BC and in the same parallels BC, EF. [I.35]

Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I.34]

And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I.34]

[But the halves of equal things are equal to one another.] [^I.37:1]

Therefore the triangle ABC is equal to the triangle DBC.

Therefore etc.

## References

[I.31]: /elem.1.31 “Book 1 - Proposition 31” [I.34]: /elem.1.34 “Book 1 - Proposition 34” [I.35]: /elem.1.35 “Book 1 - Proposition 35”

## Footnotes

[^I.37:1]: But the halves of equal things are equal to one another

Here and in the next proposition Heiberg brackets the words <quote>But the halves of equal things are equal to one another</quote> on the ground that, since the <title>Common Notion</title> which asserted this fact was interpolated at a very early date (before the time of Theon), it is probable that the words here were interpolated at the same time. Cf. note above (p. 224) on the interpolated <title>Common Notion</title>.