Properties of the Circle. 399 



EF, CF, &c. are all equal. Therefore, the sum of the areas 



of all the triangles EHF, FHC, Sic. will be equal to ^1* x 



M 



Hl + HK-f&c. = area of the whole figure. 



In like manner, the sum of all the triangles EOD, DOA, &c. 



DE 



whose vertices are at the centre O, is equal to x OG+OJN-f- 



&c. 



* ; since OG, ON, &;c. are respectively perpendicular to the 



sides ED, DA, &c. But since OG, ON, Uc. are all radii of the 

 circle; their sum is equal to the radius, multiplied by the num- 

 ber of the sides of the figure. Therefore, the sum of all the 



DE 

 triangles EOD, DOA, Stc. = x Radius of the circle, x num- 



2 

 ber of sides of the figure, = Area of the whole figure. But it 



EF 



was before shown, that — X sum of all the perpendiculars from 



DE EF 



H = Area of the whole figure. Therefore, (since — = — ,) 



the sum of all the perpendiculars from H = Radius of the cir- 

 cle, multiplied into the number of the sides of the figure. Q. 

 E. D. 



JVote. If n denote the number of the sides of the figure ; 

 and R, the radius of the circle j then will the sum of the per- 

 pendiculars be equal to nR. 



Lemma IV, 



IF the circumference of any circle be divided into any 

 number of equal parts ; and if from any point in the cir- 

 cumference, right lines be dravv^n to all the points of di- 

 vision ; the sum of the squares of all these lines will be 

 equal to tw^ice the square of the radius of the circle, 

 multiplied by the number of the parts, into which the 

 circumference is divided. Thus, if n denote the num- 

 ber of the parts ; and R, the radius of the circle ; the 

 sum of the squares of the aforesaid lines, will be equal 

 to 2nV.\ 



BEMOJ^STRATIOK. 



Let O (Plate III. Fig. 4) be any circle, the circumference of 

 which is divided into any number of equal parts, at the points 



