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

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

From :
 * $\angle EBC = \angle ECB$

Therefore by :
 * $\angle AEC = 2 \cdot \angle ECB$

From :
 * $EC = CD$

So from :
 * $\angle CED = \angle CDE$

Therefore by :
 * $\angle ECB = 2 \cdot \angle EDC$

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 :
 * $\angle BED = \angle ECB$

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

Therefore from :
 * $BD : BE = EB : BC$

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.