D是△ABC的边AC上一带点,CD=2AD,AE⊥BC交BC于E,若BD=8,sin角CBD=0.75,求AE的长

问题描述:

D是△ABC的边AC上一带点,CD=2AD,AE⊥BC交BC于E,若BD=8,sin角CBD=0.75,求AE的长

过D作DF⊥BC于F点
sin角CBD=DF/BD=0.75,所以:DF=BDsin角CBD=8*0.75=6
而AE⊥BC,知DF//AE,△CDF∽△CAE
所以:DF/AE=CD/AC,可知:AE=AC*DF/CD=6*3/2=9