Question

在△ABc中,已知AB=3,AC=6,BC=7,AD是角BAC的平分线,求证(1)DC=2BD(2)求向量AB*向量DC

Answer 1

1

AD是∠BAC的平分线,则：|AB|/|AC|=|BD|/|DC|,即：|BD|/|DC|=1/2

即：|DC|=2|BD|,故：DC=2BD

2

cosA=(b^2+c^2-a^2)/(2bc)=(36+9-49)/36=-1/9,故：AB·AC=|AB|*|AC|*cosA

=bccosA=-18/9=-2,故：AB·DC=2AB·BC/3=2AB·(AC-AB)/3=(2/3)(AB·AC-|AB|^2)

=(2/3)(-2-9)=-22/3