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 halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DBC.

Therefore, etc. Q.E.D.