400^ Demonstrations of Stexvart-s 



E, D, &;c. Take A, any point in the circumference, and; draw 

 AE, AD, &c. to all the points of division. Draw the diameter 

 AB ; and from all the points E, D, &ic. draw EM, DQ, Uc. per- 

 pendicular to AB. At the points E, D, &c. draw also tangents 

 to the circle ; and from A, draw AS, AP, &;c. perpendicular to 

 all the tangents. 



Since tiie angles at S and M are both right angles ; and the 

 Z-AES — /.AETvI ; and AE is opposite the angles at S and M ; 

 therefore AS = AM. In like manner, AP = A(4,- And in gen- 

 eral, the perpendicular from A, to any tangent, is equal to the 

 part of AB, intercepted between A, and the point where the 

 perpendicular, let fall from the point of contact of the tangent, 

 intersects AB. 



Now AE2 =AB-AM ; and AD^ =AB-Aq; and so of all the 

 lines drawn from A to the points of division. Therefore,, 



AE + AD+ &c. = ABAlvr-rAQ + &a = ABAS~+'AF+~&ic". 



But (Lemma 3d.) AS-f AP+ &.c.= ?zR ; since, if the tangents 

 be extended, they will meet, and form a regular figure, circumr- 

 scribed about the circle, in all cases, except when the circum- 

 ference is divided into only two equal parts ; in which case,,, 

 the tangents being parallel, the sum of the perpendiculars will 

 be manifestly equal to nil. Therefore, since AB = 2R ; 



AB- ASl-~iU^T&ic". = AE + AD +. &c. = 2«rI Q. E. D. 



Cor. Hence, if the arches AE, AED, he. be multiplied by 

 any number mL.n; the sum of the squares of all the lines drawn 

 from A, to the points where the multiple arches divide the cir- 



2 3 2 



cumference, will be equal to AE4-AD+ &c. =2nR. For, 

 (Lemma 1st.) the multiple arches, mAE, mAED, &c. divide the 

 circumference into equal parts ; the number of which, if m and 

 n are prime to each other, is equal to the number of the part& 

 into which the circumference was at first divided. But if m and 



n be not prime to each other, let the fraction _, when reduced. 



n 



to its lowest terms, by dividing m and n by t, its greatest com- 



k 

 mon measure, be denoted by — . Then (Lemma; M.) the arch- 



, . S , 

 es mAE, wiAED, Sic. divide the circumference into as many 

 equal parts, as there are units in g ; and as many arches termi*> 

 riate in each, point of division, as there are units in t. 



Therefore, when m and n are prime to each other, the sum 

 of the squares of all the lines drawn from A, to the points 

 where the arches mAE, wAED, Sic. terminate, will be equal 

 to 2nR^ 



