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察柏尔(chapple)定理叙述如下: 若△ABC的外接圆半径为R,内切圆半径为r,两圆的圆心距为d,则有d~2=R~2-2Rγ。证明:如图1, 连结OI并延长交外接圆于K、L。连CI并延长交外接圆于M,由相交弦定理得:KI·IL=CI·IM,又∵KI·IL=(R+d)·(R-d)=k~2-d~2,∴CI·IM=R~2-d~2 ①下面将全力以赴地寻找CI、IM与R、r的关系,为此作IP⊥BC交BC于P,则IP=r·而得CI=IP/(sin∠ICP)=r/(sin(C/2)) ②连BI,BM,在△BIM中如能证得IM=BM,则问题就会迎刃而解。因为在△CBM中能利用
Chapple’s theorem is described as follows: If ΔABC has a radius of circumscribed circle R and radius of inscribed circle is r, and the circle center distance of two circles is d, then d~2=R~2-2Rγ. Proof: As shown in Figure 1, link OI and extend the circle of circumcircle to K and L. Connect CI and lengthen the circumscribed circle to M, obtained by the intersecting string theorem: KI·IL=CI·IM, and ∵KIL=(R+d)·(Rd)=k~2-d~2,∴CI ·IM=R~2-d~2 1 The following will go all out to find the relationship between CI, IM and R, r. For this purpose, make IP ⊥ BC to submit BC to P, then IP=r· and get CI=IP/ (sin∠ICP)=r/(sin(C/2)) 2 Even BI, BM, if you can get IM=BM in △BIM, the problem will be solved. Because it can be used in △CBM