做三角形ABC的外接圆,I为三角形内心,连接AI延长与BC交于E,于圆交于D,求证ID=BD

1、<BID=<IBA+<BAI( 外角等于不相邻二内角和）,I是内心,即是角平分线的交点,BI平分<B,AI平分<A,<BID=(<A+<B)/2,<IBD=<IBE+<EBD,<EBD=<A/2(同弧圆周角相等）,<IBD=<BID,三角形DBI是等腰三角形,所以ID=BD.2、连结CD,<ECD=<DAB(同弧圆周角）,<BAD=<DAC,<ECD=<DAC,<EDC=<CDA(公用角）,△CED∽△ACD,CD/AD=DE/CD,CD^2=AD*DE,AE是角A的平分线,平分角A所对弧BC,BD弧=CD弧,BD=CD,由上题可知DB=ID,ID^2=DE*AD,6^2=y*x,xy=36,从外心O作AD垂线连结OD,在锐角和直角时DE<=DF<R,(半径是直角三角形的斜边）y<=5,x>=36/5,小于直径,10>x>=36/5, 当A角是钝角时,10>x>5.