Parallelograms About Diameter are Similar

Proof
Let $\Box ABCD$ be a parallelogram and $AC$ one of its diameters.

Let $EG, HK$ be parallelograms about $AC$.


 * Euclid-VI-24.png

We have that $EF \parallel CD$.

So from Parallel Transversal Theorem:
 * $BE : EA = CF : FA$

Again, we have that:
 * $FG \parallel CD$

So from Parallel Transversal Theorem:
 * $CF : FA = DG : GA$

From Equality of Ratios is Transitive:
 * $BE : EA = DG : GA$

So from Magnitudes Proportional Separated are Proportional Compounded:
 * $BA : AE = DA : AG$

From Proportional Magnitudes are Proportional Alternately:
 * $BA : AD = EA : AG$

Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, the sides about the common angle $\angle BAD$ are proportional.

We have that $GF \parallel DC$, $\angle AFG = \angle DCA$ and $\angle DAC$ is common to $\triangle ADC$ and $\triangle AGF$.

Therefore $\triangle ADC$ is equiangular with $\triangle AGF$.

For the same reason, $\triangle ACB$ is equiangular with $\triangle AFE$.

Thus the whole parallelogram $\Box ABCD$ is equiangular with the parallelogram $\Box EG$.

Therefore:
 * $AD : DC = AG : GF$
 * $DC : CA = GF : FA$
 * $AC : CB = AF : FE$
 * $CB : BA = FE : FA$

Since we also have:
 * $DC : CA = GF : FA$
 * $AC : CB = AF : FE$

it follows from Equality of Ratios Ex Aequali that:
 * $DC : CB = GF: FE$

Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, sides about the equal angles are proportional.

Therefore from, $\Box ABCD$ is similar to $\Box EG$.

By the same argument, $\Box ABCD$ is similar to $\Box KH$.

Therefore by Similarity of Polygons is Equivalence Relation‎, $\Box EG$ is similar to $\Box HK$.