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 $$CD = CF = CE$$.


 * LuneOfHippocrates.png

Consider the lune $$DFEB$$.

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

Proof

 * LuneOfHippocratesProof.png

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 $$\left({\sqrt 2}\right)^2$$ 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 $$AEGC$$.