Quadrature of Lunes

The General Problem: Find a square that has the same area as that enclosed by a given lunar region.

Special Case: An Isosceles Right Triangle. [Hippocrates/ Eudemus-Aristotle]

Suppose the outer arc is a semicircle with diameter the base of isosceles right triangle and the inner arc is similar to the arc cut by the isosceles right triangle in the outer circumference.
Then the area of the lune is the area of the right triangle.

Reason: The area of the two smaller circular regions between the outer arc and the triangle is equal to the area of the larger circular region between the triangle and the inner arc.[This combines the similarity of the circular regions with the Pythagorean theorem and the dissection of the triangle.]

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Alternative Figure and Proof [Hippocrates?/Alexander-Simplicius]

• L+ S = C1 ;
• 2S + 2T = C2;
• 2C1 = C2 [Pythagorus] ;
• 2S + 2T = 2C1 ;
• S + T = C1 ;
• Therefore, L = T.

Reference: Calinger, Classics of Mathematics (1995) pp. 59-62.