Similar Parallelogram on Half a Straight Line

Theorem

In the words of Euclid:

Of all the parallelograms applied to the same straight line and deficient by parallelogramic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

(The Elements: Book $\text{VI}$: Proposition $27$)

Proof

Let $AB$ be a straight line and let it be bisected at $C$.

Let the parallelogram $\Box AD$ be applied on $AB$ deficient by the parallelogramic figure $\Box DB$ described on $AC$.

We need to show that of all the parallelograms applied to $AB$ and deficient by parallelogramic figures similar and similarly situated to $\Box DB$, $\Box AD$ is the greatest.

Let $\Box AF$ be applied to $AB$ deficient by $\Box FB$ similar and similarly situated to $\Box DB$.

We have that $\Box DB$ is similar to $\Box FB$.

So from Parallelogram Similar and in Same Angle has Same Diameter they are about the same diameter.

Let that diameter $DB$ be drawn, and let that figure be described.

From Complements of Parallelograms are Equal:

$\Box CF = \Box FE$

As $\Box FB$ is common:

$\Box CH = \Box KE$

But from Parallelograms with Equal Base and Same Height have Equal Area:

$\Box CH = \Box CG$ since $AC = CB$

Therefore:

$\Box GC = \Box EK$

Add $\Box CF$ to each.

Therefore $\Box AF$ is equal in area to the gnomon $LMN$.

So $\Box DB$, that is $\Box AD$, is greater than $\Box AF$.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $27$ of Book $\text{VI}$ of Euclid's The Elements.

Sources

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