Parallelogram Similar and in Same Angle has Same Diameter

Theorem

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

Proof
From the parallelogram $\Box ABCD$ let there be taken away the parallelogram $\Box AF$ similar and similarly situated to the whole and having a common angle with it.

We need to show that $\Box ABCD$ is about the same diameter with $\Box AF$.


 * Euclid-VI-26.png

Suppose it is not, and that $AHC$ is the diameter of $\Box ABCD$.

Let $GF$ be produced to $H$ and from Construction of a Parallel let $HK$ be drawn parallel to $AD$ and $BC$.

From Parallelograms About Diameter are Similar, $GA : AB = GA : AK$.

But also because of the similarity of $\Box ABCD$ and $\Box EG$, $DA : AB = GA : AE$.

So from Equality of Ratios is Transitive $GA : AK = GA : AE$.

Therefore from Magnitudes with Same Ratios are Equal $AK = AE$.

Hence the result.