Parallelograms which are on equal bases and in the same parallels are equal to one another.
[Desmos graphs can be found here]
Consider the parallelograms ABCD and EFGH which are on equal bases BC and FG and in the same parallels AH and BG as shown.
We will prove that the parallelogram ABCD is equal to the parallelogram EFGH.
Let BE and CH be joined.
Since BC is equal to FG (hypothesized equal bases) and FG is equal to EH by Proposition 34 (in a parallelogram, opposite sides and opposite angles are equal), then BC is equal to EH.
BC is also parallel to EH, and they are joined by BE and CH.
By Proposition 33 (straight lines that join equal and parallel lines are themselves also parallel and equal), BE and CH are equal and parallel.
Therefore EBCH is a parallelogram by Proposition 34.
The parallelogram EBCH is equal to the parallelogram ABCD by Proposition 35 (parallelograms on the same base and in the same parallels are equal to one another) because they have the same base BC and are in the same parallels BC and AH.
Also, by Proposition 35, the parallelogram EBCH is equal to the parallelogram EFGH as they share the base EH and are in the same parallels EH and BG.
Since both ABCD and EFGH are equal to EBCH, they are equal to each other. Therefore the parallelogram ABCD is equal to the parallelogram EFGH.
Q.E.D.
In the next proposition we will prove a similar result for triangles – George