58 SCIENCE IN GRECO-ROMAN ANTIQUITY 



a 



x 



x 



y 



y. 



b 



Then we have x 2 = ay ; y 2 = xb, hence # 4 = a 2 xb 

 and x z = a 2 b. Now if we put b — 2a we obtain 

 x z = 2a 3 , which is the solution required. 



The quadrature of the circle is, as we know, an 

 insoluble geometrical problem. In attempting to 

 solve it, Hippocrates was led to several interesting 

 discoveries on lunes. He found, for example, that 

 the lune AECD (Fig. 6) is equal to half the right- 



angled triangle ACB. In order to prove it, it is 

 sufficient to notice that the semi-circle constructed on 

 the hypotenuse BC is equal in area to the two semi- 

 circles constructed on the sides BA and AC which, by 

 hypothesis, are equal. If we take away the common 

 parts of the semi-circles (small and large) we obtain the 

 required equality. 1 Having thus demonstrated that 

 a surface bounded by curvi-linear elements is equal 

 to a surface limited by straight lines, Hippocrates 

 thought it was possible to find a square equal to a 



1 23 Rouse Ball, History of Mathematics, I, p. 42. 



