In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.

Let *ABCD* be a parallelogram, and *AC* its diameter, and let *EG* and *HK* be parallelograms about *AC.*

I say that each of the parallelograms *EG* and *HK* is similar both to the whole *ABCD* and to the other.

For, since *EF* is parallel to a side *BC* of the triangle *ABC,* proportionally, *BE* is to *EA* as *CF* is to *FA.*

Again, since *FG* is parallel to a side *CD* of the triangle *ACD,* proportionally, *CF* is to *FA* as *DG* is to *GA.*

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

And, since *GF* is parallel to *DC,* the angle *AFG* equals the angle *ACD,* and the angle *DAC* is common to the two triangles *ADC* and *AGF,* therefore the triangle *ADC* is equiangular with the triangle *AGF.*

For the same reason the triangle *ACB* is also equiangular with the triangle *AFE,* and the whole parallelogram *ABCD* is equiangular with the parallelogram *EG.*

Therefore, proportionally, *AD* is to *DC* as *AG* is to *GF, DC* is to *CA* as *GF* is to *FA, AC* is to *CB* as *AF* is to *FE,* and *CB* is to *BA* as *FE* is to *EA.*

And, since it was proved that *DC* is to *CA* as *GF* is to *FA,* and *AC* is to *CB* as *AF* is to *FE,* therefore, *ex aequali, DC* is to *CB* as *GF* is to *FE.*

Therefore in the parallelograms *ABCD* and *EG* the sides about the equal angles are proportional. Therefore the parallelogram *ABCD* is similar to the parallelogram *EG.*

For the same reason the parallelogram *ABCD* is also similar to the parallelogram *KH.* Therefore each of the parallelograms *EG* and *HK* is similar to *ABCD.*

But figures similar to the same rectilinear figure are also similar to one another, therefore the parallelogram *EG* is also similar to the parallelogram *HK.*

Therefore, *in any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.*

Q.E.D.