Rectangles Contained by Three Proportional Straight Lines

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Theorem

In the words of Euclid:

If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes is equal to the square on the mean, the three straight lines will be proportional.

(The Elements: Book $\text{VI}$: Proposition $17$)


Proof

Let the three straight lines $A, B, C$ be proportional, that is:

$A : B = B : C$

Then we need to show that the rectangle contained by $A$ and $C$ equals the square on $B$.

Euclid-VI-17.png

Let $D = B$.

Then $A : B = D : C$

By Rectangles Contained by Proportional Straight Lines, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$.

But as $B = D$, the rectangle contained by $B$ and $D$ equals the square on $B$.

So the rectangle contained by $A$ and $C$ equals the square on $B$.

$\Box$

Now let the rectangle contained by $A$ and $C$ be equal to the square on $B$.

Using the same construction, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$ because $B = D$.

So by Rectangles Contained by Proportional Straight Lines:

$A : B = D : C$

But as $B = D$:

$A : B = B : C$

$\blacksquare$


Historical Note

This proof is Proposition $17$ of Book $\text{VI}$ of Euclid's The Elements.
This is, of course, just a special case of Proposition $16$ of Book $\text{VI} $: Rectangles Contained by Proportional Straight Lines.


Sources

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