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

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

In the words of Hypsicles of Alexandria:

If an equilateral and equiangular pentagon be inscribed in a circle, the sum of the squares on the straight line subtending two sides and on the side of the pentagon is five times the square on the radius.

(The Elements: Book $\text{XIV}$: Proposition $2$ : Lemma)

Proof

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:

\(\ds BE\)

\(=\)

\(\ds 2 \cdot ED\)

\(\text {(1)}: \quad\)

\(\ds \therefore \ \ \)

\(\ds BE^2\)

\(=\)

\(\ds 4 \cdot ED^2\)

Also:

\(\ds \angle BAE\)

\(=\)

\(\ds 90^\circ\)

Thales' Theorem

\(\ds \therefore \ \ \)

\(\ds AB^2 + AE^2\)

\(=\)

\(\ds BE^2\)

Pythagoras's Theorem

\(\ds \therefore \ \ \)

\(\ds AB^2 + AE^2 + ED^2\)

\(=\)

\(\ds 5 \cdot ED^2\)

from $(1)$

But from Proposition $10$ of Book $\text{XIII} $: Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle:

$AC^2 = DE^2 + EA^2$

Therefore:

$BA^2 + AC^2 = 5 \cdot DE^2$

$\blacksquare$

Historical Note

This proof is Proposition $2$ of Book $\text{XIV}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): The So-Called Book $\text{XIV}$, by Hypsicles