(必须用正弦定理证明）如图 在△ABC中 BD是∠ABC的角平分线,求证AB/BC=AD/DC

三角形ABD中,sin角ABD/AD=sin角ADB/AB,即AB*sin角ABD=AD*sin角ADB三角形CBD中,sin角CBD/DC=sin角CDB/BC,即BC*sin角CBD=CD*sin角CDB两式相除,得（AB*sin角ABD）/（BC*sin角CBD）=（AD*sin角ADB）/（CD*sin角CDB）即（AB/BC)*(sin角ABD/sin角CBD)=（AD/CD)*(sin角ADB/sin角CDB）因为角ABD=角CBD,因此sin角ABD/sin角CBD=1角ADB+角CDB=180度,sin角ADB/sin角CDB=sin角（180度-角CDB）/sin角CDB=1因此AB/BC=AD/CD