如图,CA、CB为圆o的切线,切点分别为A、B.直径延长AD与CB的延长线交于点E.AB、CO交于点M,连接OB. (1)求证:角ABO=1/2角ACB (2)若sin角EAB=根号10/10,CB=12,求圆o的半径及BE/AE的值.
1）
因为AC & BC切圆,因此角CAO = 角CBO = 90度
角ACB + 角AOB = 180度 => 角ACB = 角BOE
三角形AOB,因为OA = OB,因此角ABO = 角BAO
角ABO + 角BAO = 角BOE
=> 2角ABO = 角ACB
2)
因为角CAO = 角CBO = 90度,OA = OB,共用OC,因此三角形AOC全等於三角形BOC
角ACO = 角BCO = 角ABC / 2 = 角ABO = 角BAO
BO = 12 * (根号10 / 根号90) = 12 / 3 = 4
因为角BOE = 2角BAO
sin(角BOE) = sin(2角BAO) = 2 sin(2角BAO) * cos(角BAO) = 2 * (根号10/10) * (根号90/10) = 3/5 = => BE :BO :OE = 3 :4 :5
BE / AE = BE / (AO + OE) = BE/(BO + OE) = 3 / (4 + 5) = 1/3
