Similar Parallelogram on Half a Straight Line

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.


 * Euclid-VI-27.png

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.