If from a parallelogram there is taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, then it is about the same diameter with the whole.

From the parallelogram *ABCD* let there be taken away the parallelogram *AF* similar and similarly situated to *ABCD,* and having the angle *DAB* common with it.

I say that *ABCD* is about the same diameter with *AF.*

For suppose it is not, but, if possible, let *AHC* be the diameter. Produce *GF* and carry it through to *H.* Draw *HK* through *H* parallel to either of the straight lines *AD* or *BC.*

Since, then, *ABCD* is about the same diameter with *KG,* therefore *DA* is to *AB* as *GA* is to *AK.*

But also, since *ABCD* and *EG* are similar, therefore *DA* is to *AB* as *GA* is to *AE.* Therefore *GA* is to *AK* as *GA* is to *AE.*

Therefore *GA* has the same ratio to each of the straight lines *AK* and *AE.*

Therefore *AE* equals *AK* the less equals the greater, which is impossible.

Therefore *ABCD* cannot fail to be about the same diameter with *AF.* Therefore the parallelogram *ABCD* is about the same diameter with the parallelogram *AF.*

Therefore, *if from a parallelogram there is taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, then it is about the same diameter with the whole.*

Q.E.D.