Rectangles Contained by Proportional Straight Lines

Theorem

 * ''If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes is equal to the rectangle contained by the means, the four lines will be proportional.

Note: in the above, equal is to be taken to mean of equal area.

Proof
Let the four straight lines $AB, CD, E, F$ be proportional, that is, $AB : CD = E : F$.

What we need to show is that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.


 * Euclid-VI-16.png

Let $AG, CH$ be drawn perpendicular to $AB$ and $CD$.

Let $AG = F$, $CH = E$.

Complete the parallelograms $BG$ and $DH$.

We have that $AB : CD = E : F$, while $E = CH$ and $F = AG$.

So in $\Box BG$ and $\Box DH$ the sides about the equal angles are reciprocally proportional.

But from Sides of Equiangular Parallelograms are Reciprocally Proportional‎, $\Box BG = \Box DH$ (in area).

We also have that:
 * $\Box BG$ is the rectangle contained by $AB$ and $F$
 * $\Box DH$ is the rectangle contained by $CD$ and $E$

Hence the result.

Now suppose that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.

We use the same construction, and note that $\Box BG = \Box DH$ (in area).

But they are equiangular, as all angles are equal to a right angle.

So from Sides of Equiangular Parallelograms are Reciprocally Proportional‎ $AB : CD = CH : AG$.

But $E = CH$ and $F = AG$.

So $AB : CD = E : F$.