"He commenced by finding the area of a lune contained between a semicircle and a quadrilateral arc standing on the same chord... as follows. Let ABC be an isosceles right-angled triangle inscribed in the semicircle ABOC, whose centre is O. On AB and AC as diameters describe semicircles as in the figure. Then, since by Euc. I, 47,sq. on\,BC = sq. on\,AC + sq. on\,AB,therefore, by Euc. XII, 2,area\;\frac{1}{2} \bigodot on\,BC = area\;\frac{1}{2} \bigodot on\,AC + area\;\frac{1}{2} \bigodot on\,ABTake away the common parts.\therefore area\,\triangle ABC = sum\;of\;areas\;of\;lunes\;AECD\;and\;AFBG.Hence the area if the lune AECD is equal to half that of the triangle ABC."
January 1, 1970
https://en.wikiquote.org/wiki/Hippocrates_of_Chios