Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio

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Theorem

In the words of Euclid:

If the side of the hexagon and decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.

(The Elements: Book $\text{XIII}$: Proposition $9$)


Proof

Euclid-XIII-9.png

Let $ABC$ be a circle.

Let $BC$ be the side of a regular decagon which has been inscribed within $ABC$.

Let $BC$ be produced to $D$, where $CD$ is equal to the side of a regular hexagon which can be inscribed within $ABC$.

It is to be demonstrated that the straight line $BCD$ is cut in extreme and mean ratio at the point $C$, and that $CD$ is the greater segment.


Let $E$ be the center of $ABC$.

Let $EB, EC, ED$ be joined.

Let $BE$ be produced to $A$.

We have that $BC$ is the side of a regular decagon.

Therefore the arc $ACB$ is five times the arc $CB$.

Therefore the arc $AC$ is five times the arc $CB$.

But from Proposition $33$ of Book $\text{VI} $: Angles in Circles have Same Ratio as Arcs:

$\angle AEC = 4 \cdot \angle CEB$

From Proposition $5$ of Book $\text{I} $: Isosceles Triangle has Two Equal Angles:

$\angle EBC = \angle ECB$

Therefore by Proposition $32$ of Book $\text{I} $: Sum of Angles of Triangle equals Two Right Angles:

$\angle AEC = 2 \cdot \angle ECB$

From Porism to Proposition $15$ of Book $\text{IV} $: Inscribing Regular Hexagon in Circle:

$EC = CD$

So from Proposition $5$ of Book $\text{I} $: Isosceles Triangle has Two Equal Angles:

$\angle CED = \angle CDE$

Therefore by Proposition $32$ of Book $\text{I} $: Sum of Angles of Triangle equals Two Right Angles:

$\angle ECB = 2 \cdot \angle EDC$

But it has been proved that:

$\angle AEC = 2 \cdot \angle ECB$

Therefore:

$\angle AEC = 4 \cdot \angle EDC$

But it has been proved that:

$\angle AEC = 4 \cdot \angle BEC$

Therefore:

$\angle EDC = \angle BEC$

But $\angle EBD$ is common to $\triangle BEC$ and $\triangle BED$.

Therefore by Proposition $32$ of Book $\text{I} $: Sum of Angles of Triangle equals Two Right Angles:

$\angle BED = \angle ECB$

Therefore $\triangle EBD$ is equiangular with $\triangle EBC$.

Therefore from Proposition $4$ of Book $\text{VI} $: Equiangular Triangles are Similar:

$BD : BE = EB : BC$

But $EB = CD$

Therefore:

$BD : DC = DC : CB$

and:

$BD > DC$

Therefore:

$DC > CB$

Therefore $BD$ is cut in extreme and mean ratio at the point $C$, and $CD$ is the greater segment.

$\blacksquare$


Historical Note

This proof is Proposition $9$ of Book $\text{XIII}$ of Euclid's The Elements.


Sources

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