Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere/Lemma

Proof

 * Euclid-XIV-2-Lemma.png

Let $ABC$ be a circle.

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

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

Let $DF$ be drawn from $D$ perpendicular to $AC$, and produce it both ways to meet $ABC$ in $B$ and $E$.

Let $AB$ and $AE$ be joined.

It is to be demonstrated that:
 * $BA^2 + AC^2 = 5 \cdot DE^2$

We have that:

Also:

But from :
 * $AC^2 = DE^2 + EA^2$

Therefore:
 * $BA^2 + AC^2 = 5 \cdot DE^2$