Proposition 37
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 line AD and line BC.
I say that the triangle ABC equals the triangle BCD.

Produce line AD in both directions to point E and point F. Draw line BE through point B parallel to line AC, and draw CF through C parallel to line BD.
Then each of the figures polygon EBCA and polygon DBCF is a parallelogram, and they are equal, for they are on the same base BC and in the same parallels line BC and line EF.

Moreover the triangle ABC is half of the parallelogram EBCA, for the diameter line AB bisects it. And the triangle BCD is half of the parallelogram DBCF, for the diameter line CD bisects it.

Therefore the triangle ABC equals the triangle BCD.
Therefore triangles which are on the same base and in the same parallels equal one another.