Triangles which are on the same base and in the same parallels are equal to one another.
[Desmos graphs can be found here]
Let ABC and DBC be triangles on the same base BC and in the same parallels AD and BC.
We will prove that triangles ABC and DBC are equal.
Project the line AD in both directions to points E and F such that BE is parallel to CA and CF is parallel to BA.
Proposition 31 covered how to construct a line through a given point that is parallel to a given line.
(BE passes through B and is parallel to CA. CF passes through C and is parallel to BA.)
EBCA and DBCF are parallelograms. The two parallelograms share the same base (BC) and are in the same parallels (BC and EF), so by Proposition 35 they are equal.
By Proposition 34 (the diameter of a parallelogram bisects its area) triangle ABC is half of the parallelogram EBCA because its diameter AB bisects the parallelogram.
Again by Proposition 34, since DC is the diameter of the parallelogram DBCF, triangle DBC is half of the parallelogram DBCF.
Since the triangles ABC and DBC are each half of equal parallelograms EBCA and DBCF, the two triangles ABC and DBC are therefore equal to one another.
Q.E.D.
The next proposition extends Proposition 37 from “the same base” to “equal bases” – George