Perpendicular from Center of Circle to Side of Inscribed Pentagon

Proof

 * Euclid-XIV-1.png

Let $ABC$ be a circle.

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

Let $D$ be the center of the circle $ABC$.

Let $DE$ be drawn from $D$ perpendicular to $BC$, and produce it both ways to meet $ABC$ in $A$ and $F$.

It is to be demonstrated that $DE$ is half the sum of:
 * the side of a regular hexagon which has been inscribed within $ABC$

and:
 * the side of a regular decagon which has been inscribed within $ABC$

Let $DC$ and $CF$ be joined.

Let $G$ be the point on $AE$ such that $GE = EF$.

Let $GC$ be joined.

We have that:
 * the circumference of $ABC$ is $5$ times $\smile BFC$

and:
 * half the circumference of $ABC$ is $\smile ACF$

where $\smile BFC$ and $\smile ACF$ are used to denote the arc $BFC$ and arc $ACF$.

while:
 * $\smile FC$ is half of $\smile BFC$.

Thus:

Also:

But from :
 * $DE$ is the side of a regular hexagon which has been inscribed within $ABC$

and from the construction:
 * $FC$ is the side of a regular decagon which has been inscribed within $ABC$.

Hence the result.