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

Let *ABC* and *DBC* be triangles on the same base *BC* and in the same parallels *AD* and *BC.*

I say that the triangle *ABC* equals the triangle *DBC.*

Produce *AD* in both directions to *E* and *F.* Draw *BE* through *B* parallel to *CA,* and draw *CF* through *C* parallel to *BD.*

Then each of the figures *EBCA* and *DBCF* is a parallelogram, and they are equal, for they are on the same base *BC* and in the same parallels *BC* and *EF.*

Moreover the triangle *ABC* is half of the parallelogram *EBCA,* for the diameter *AB* bisects it. And the triangle *DBC* is half of the parallelogram *DBCF,* for the diameter *DC* bisects it.

Therefore the triangle *ABC* equals the triangle *DBC.*

Therefore *triangles which are on the same base and in the same parallels equal one another.*

Q.E.D.

The justification of the last conclusion is missing. From the statement that the doubles of two magnitudes are equal, we want to conclude that the magnitudes themselves are equal. Although Euclid included no such common notion, others inserted it later. See the commentary on Common Notions for a proof of this halving principle based on other properties of magnitudes.