# Lune of Hippocrates

## Theorem

Take the circle whose center is $A$ and whose radius is $AB = AC = AD = AE$.

Let $C$ be the center of a circle whose radius is $CD = CF = CE$.

Consider the lune $DFEB$.

Its area is equal to that of the square $AEGC$.

## Proof

The chords $DB$ and $EB$ are tangent to the arc $DFE$. They divide the lune into three regions: yellow, green and blue.

From Pythagoras's Theorem, $CD = \sqrt 2 AD$.

The green and blue areas are of equal area as each subtend a right angle.

The orange area also subtends a right angle.

So the area of the orange area is $\paren {\sqrt 2}^2$ the area of either the green or blue areas.

That is, the orange area equals the sum of the green and blue areas together.

Thus the area of the lune $DFEB$ is equal to the area of $\triangle DEB$.

It is a simple matter then to show that the area of $\triangle DEB$ is the same as the area of the square $AEGC$.

$\blacksquare$

## Source of Name

This entry was named for Hippocrates of Chios.

## Historical Note

The Lune of Hippocrates was the first time a precise measure was made of an area bounded by curved lines.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.4$: The Lunes of Hippocrates