题目如图,点DDD是△ABC\triangle ABC△ABC边BCBCBC上一点,连接ADADAD,使∠BAD=∠C\angle BAD=\angle C∠BAD=∠C,则下列结论正确的是( )A.AC:BC=AD:BDAC:BC=AD:BDAC:BC=AD:BDB.AB2=CD⋅BCAB^{2}=CD\cdot BCAB2=CD⋅BCC.AC:BC=AB:ADAC:BC=AB:ADAC:BC=AB:ADD.AB2=BD⋅BCAB^{2}=BD\cdot BCAB2=BD⋅BC知识点:相似三角形的判定I章节:图形的变化 / 图形的相似