Parallelogram Similar and in Same Angle has Same Diameter

Theorem

In the words of Euclid:

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.

(The Elements: Book $\text{VI}$: Proposition $26$)

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 $\Box AF$.

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

Let $GF$ be produced to $H$.

From Construction of Parallel Line let $HK$ be drawn parallel to $AD$ and $BC$.

From Parallelograms About Diameter are Similar:

$DA : AB = GA : AK$

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.

$\blacksquare$

Historical Note

This proof is Proposition $26$ of Book $\text{VI}$ of Euclid's The Elements.
It is the converse of Proposition $24$: Parallelograms About Diameter are Similar.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions