Lune of Hippocrates - Proof

Proof

Hippocrates' result can be proved as follows: The center of the circle on which the arc AEB lies is the point D, which is the midpoint of the hypotenuse of the isosceles right triangle ABO. Therefore the diameter AC of the larger circle ABC is √2 times the diameter of the smaller circle on which the arc AEB lies. Consequently the smaller circle has half the area of the larger circle, and therefore the quarter circle AFBOA is equal in area to the semicircle AEBDA. Subtracting the crescent-shaped area AFBDA from the quarter circle gives triangle ABO and subtracting the same crescent from the semicircle gives the lune. Since the triangle and lune are both formed by subtracting equal areas from equal area, they are themselves equal in area.

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