Equal triangles which are on equal bases and on the same side are also in the same parallels.
[Desmos graphs can be found here]
Let ABC and CDE be equal triangles with equal bases BC and CE.
We will prove that AD is parallel to BE to show that the two triangles are in the same parallels.
Begin by joining AD.
If AD is not parallel to BE, then draw AF through point A so that AF is parallel to BE. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join FE as well.
Triangles ABC and FCE have equal bases (BC and CE) and in the same parallels (BE and AF), so by Proposition 38 (triangles on equal bases and in the same parallels are equal to one another) triangles ABC and FCE are equal to one another.
Since triangle ABC is also equal to triangle CDE, triangle FCE must also be equal to triangle CDE which is impossible. (One of those triangles must be greater than the other.)
Therefore AF is not parallel to BE.
Therefore AD is parallel to BE, and the two triangles are in the same parallels.
Q.E.D.
The next proposition will compare triangles and parallelograms with equal bases in the same parallels – George