VOL. XXI.] PHILOSOPHICAL TRANSACTIONS. 453 



the squares of their diameters are as 2 to 1 ; and in the same proportion are 

 their respective circles ; and therefore a quarter of the one equal to half the 

 other. Consequently, if from each of these be subtracted the common seg- 

 ment abd : the remaining lunula adbe on the one side, will be equal to the 

 remaining triangle abc, on the other side ; or to abk, supposing ab bisected 

 in K, that is, to half the square ok, inscribed in the lesser circle. Which is 

 commonly called, the squaring of Hippocrates's lunula, that is, the finding a 

 rectilinear figure equal to that lunula. 



This being premised, the point in hand is the squaring a given portion of 

 such lunula ; suppose ade, cut ofFby a straight line ode, drawn from the centre 

 c. Which Mr. Perks performs after this manner, viz. Drawing the straiglit 

 lines ea and eb, fig. 2, cutting the arc eb in g, and on ag a perpendicular ef, 

 which will pass to the centre c, because bisecting ag at right angles ; the right 

 lined triangle afe is equal to ade, the proposed portion of the lunula. 



Demonstration. — adb being a quadrantal arc, the angle agb will be three 

 halves of a right angle, and its conjunct angle ega, half a right angle. And 

 that angle being external to the triangle age, is equal to the two opposite inter- 

 nals GEA + eag. Thereof gea, because an angle in the semicircle aeb, is a 

 right angle, and therefore eag is half a right angle, as are also peg and fea. 

 And the three triangles afe, gfe, and gea, each of them half a square. And 

 AG to ae, as \/2 to 1, proportional to the respective radii of the two circles. 

 And the like segments adg, ae, in their respective circles, as the squares of 

 their respective radii, as 2 to 1. And therefore the semisegment afd, equal to 

 the segment ae. Consequently, one taking from the triangle as much as the 

 other adds to it, the portion of the lunula ade equal to the triangle afe. 



If the point e chance to be in k, fig. 1, the middle of the arc aeb, there 

 will be no intersection at g, the points g, b, coinciding, but without any dis- 

 turbance to the demonstration. If it fall beyond it, toward b, then g will be 

 on the other side, and what is here said of egb must be accommodated to ega, 

 which things are so obvious, as not to need any long discourse. And this is 

 applicable to other lunulas beside that of Hippocrates, if, by altering the angle 

 at F, or otherwise, we take in such a portion of the common segment abd on 

 the one side, instead of ae cut off on the other side, as the proportion of the 

 two circles requires. 



I showed this quadrature of Mr. Perks to Dr. David Gregory, our learned 

 professor of astronomy at Oxford, who gives his opinion about it, with his im- 

 provement of it, as follows : 



" The quadrature of the parts of the lunula of Hippocrates of Chios, by 

 Mr. Perks, is very elegant. I remember the like was done some years since by 



