In any parallelogram the complements of the parallelogram about the diameter are equal to one another.
[Desmos graphs can be found here]
Let ABCD be a parallelogram with diameter AC.
About AC let AEKH and KGCF be parallelograms.
Call BEKG and DHKF the complements.
We will prove that the parallelograms BEKG and DHKF are equal to one another.
Since ABCD is a parallelogram and AC its diameter, Proposition 34 (the diameter of a parallelogram bisects it) tells us that triangle ABC is equal to the triangle ACD.
The parallelogram AEKH has a diameter AK, so again by Proposition 34 triangle AEK is equal to the traingle AHK.
KC is a diameter of the parallelogram KGCF, so by Proposition 34 triangle KGC is equal to triangle KFC.
Triangle ABC is composed of triangle AEK, triangle KGC, and parallelogram BEKG.
Triangle ADC is composed of triangle AHK, triangle KFC, and parallelogram DHKF.
Triangles ABC and ADC are equal to each other, and removing the two sets of equal triangles from each (AEK = AHK and KGC = KFC), the complements (parallelograms BEKG and DHKF) must be equal to each other.
Therefore parallelogram BEKG is equal to parallelogram DHKF. The complements of the parallelograms about the diameter are equal to one another.
Q.E.D.
The next proposition is a construction that relies on the result in Proposition 43 – George