( 2S 0IVISION OF A CIRCLE. 



Conftruftion, 



Dlvifionofan From any point, Y, in the right line A B, draw Y Wat 

 arch of a circle r |ght angles to the fame, making it equal to the given right 

 having their ' '» ie ** i through W, parallel to Y B, draw WF, and let it cut 

 fines, or cofmes, t h e arcn a B in F, then will A F, F B, be the required arches, 

 or v. fines, in a 



8iven rati0 ' Demonjlration. 



Draw F Q perpendicular to YB, then F Q* = W Y* = 

 m x m ; by Conft. and Euc. 34. I. ; but the rectangle of the 

 verfed fines of A F and FB = F*qK (by Cor. 3. Prop. II.) ; 

 therefore this re&angle is equal to m x m, the given fquare. 

 Q. E. D. 



Proposition IV. Problem. 



To divide A F B, a given arch of a circle, (See Fig. 3.) into 

 two parts, A F, F B, fo that the rectangle of their fines may 

 be equal to a given fquare, (n x n) ? 



Conftruclian. 



To make the conftru&ion general, let AFB be greater 

 than a femicircle, join A B, and in it take Q, making A Q 

 X Q B = n x n ; alfo in A B produced take q, fo as to make 

 AqxqR=nxn; draw Q F, qfg, perpendicular to A B ; 

 then will A F, F B, or Ag, g B, or A/, /B, be the required 

 arches. 



De7iion ft ration. 



This is evident from Cor. 3. Prop. II. and the conftrucVion. 



To find the limits, bifecl A B in Z, draw alfo the radius 

 O N parallel to Z B, and make N E perpendicular to A B 

 produced ; then, if ;n x n be greater than A Z x Z B, F is 

 an imaginary point, becaufe A Q x Q B cannot exceed A Z 

 X Z B, by Euc. 5. II. Again, if n x n be greater than A E 

 X E B, the points/, g t are imaginary, becaufe Aq X q B 

 cannot exceed A E X E B, feeing E N touches the circle in 

 N, and is parallel to qg: Thefe things being premifed, it 

 will be eafily perceived, that when A F B is lei's than a femi- 

 circle, it can only be divided in one point to anfwer the con- 

 ditions of the queftion, becaufe the point N will be in the op- 

 pofite fegment; but when it exceeds a femicircle, it will ad- 

 mit of being divided into one, two, or three points, according 

 to circumftances, or even the conftru&ion may prove impof- 



fible. Q, E. D. 



Schojjum 



