404 Demonstrations of Stewart'^s 



CP "2 = P"P2 + PC" - ^^-?P—. By reduction, CP'^ =^ 



PQ, ^ 



P'P2^Pg4-PC2 ■ P Q^+2PP PC^ _ pp2.2PO_ |_ PC^-2P0 „ 



■ PQ, 2P6 

 PP^PO+PC^PO. . ,^p,,_ P P^PO+PC^PO- g^^ 



PO PO 



P'P = P'0-PO, and P"P=PO-P"0. Therefore, CP'^ = 

 <P-0-PO/ -P O+PC'-P'Q pp ,2^(PQ-P 0/ PQ +PC^ F'Q 

 PO '^ PO '^ 



Wherefore CP'^ 



(PO-PO)^PO^+2(PO-PO)"POPC^PO+PC-'PQ^ 



PU* ' 



and CP'"* 



_ (PO-F 0)^PO^+2(PO-P 0/ PO PC^P' O-t-PC^ P O - 



PQ^2 — • 



By the same mode of reasoning, it may be shewn, that BP''* = 

 ; P»Q-P0)4 PG24-2(P'0-POf -PO •PB^-P'Q + PB-^-P'Q g, 



- PQ2 • - ? 



and BP"^ = 



(PO-P Q)^'P0g+2(P0-P 0)2P0PB"P O+PB^P 0^ 



■ PO^ ' 

 and so of all the other lines drawn from the angular points to 

 V or P". 



Let 71 denote the number of the sides of the figure ; and 

 E, the radius, or PO. Then CP'^ + BP'^-j- ^^c. = 



R^-p*'^'^li)''T(P'0-HrTl^. + 2(P'0-R)^-P '0-U_x 



w 



■ pC^+vWy^c. + P-0^ -PCy PB44- .^c. But, (P'O-R)^ 



-f (P'O-R)^ + 4-c. =n(P'0-R)^ And (Lemma 4th.) PCH 

 PB^ 4- ^c.= 2nR\ And (Prop. L) PC^PB-* + <^c. = 6nR*. 

 Hence, (JP'^+BP'4+ <^c.= 



_ nW (PO-R)-t + 4nR^ (P'O - Uf • P'O - R -f eraR^* • P'O^ _ 



■"' ■ j^i ' - ~- 



n ' (P^^^)"^+ T("P^O'^"R7^'^P^O~Tr+~6R^P'02 = n • 

 p>0^^~4F05~Rr4r6F0^Tl^^F0T{H^ 



8P'02 • R2+4P'0-R3-t-6P'02 -R^ = n •P'0*+4P'0^ -R^ + R^ 



