Parallelograms that are on the same base and in the same parallels are equal to one another.
We will start with two parallelograms on the same base BC and in the same parallels BC and AF.
We will prove that the parallelograms ABCD and EBCF shown below are equal.
Here are both parallelograms on the same graph.
By Proposition 34 (in a parallelogram opposite sides and opposite angles are equal) we know that AD is equal to BC as they are opposite sides in the parallalogram ABCD.
Looking to parallelogram EBCF, Proposition 34 tells us that EF is also equal to BC.
Therefore AD is equal to EF.Adding DE to each we find that AE (AD & DE) is equal to DF (DE & EF).
Again using Proposition 34, AB is equal to DC.
So EA is equal to FD, AB is equal to DC, and the angles contained (FDC and EAB) are equal by Proposition 29 (if a straight line falls on two parallel lines, the exterior angle is equal to the interior and opposite angle).
Therefore triangle EAB is equal to the triangle FDC by Proposition 4 (side-angle-side).
Subtract the triangle DGE from both triangles EAB and FDC.
The resulting trapezoids ABGD and EGCF are equal.
Add triangle GBC to each trapezoid.
Therefore the resulting parallelograms ABCD and EBCF are equal.
Q.E.D.
In the next proposition we will extend Proposition 35 to include bases that are not the same but are equal – George