在△ABC中,AB=4,AC=3,角A的平分线AD交BC于D,且AD=2,求BC用余弦定理

Question

在△ABC中,AB=4,AC=3,角A的平分线AD交BC于D,且AD=2,求BC
用余弦定理

Answer 1

AD是平分线
|AB|/|AC|=|BD|/|DC|=2/3
即：|BD|=2|DC|/3,即：BD=2BC/5
故：AD=AB+BD=AB+2(AC-AB)/5
=3AB/5+2AC/5
cosA=(b^2+c^2-a^2)/(2bc)=(9+4-16)/12=-1/4
故：|AD|^2=(1/25)(9|AB|^2+4|AC|^2+12AB·AC)
=(1/25)(36+36+12*2*3(-1/4))
=54/25,即：|AD|=3√6/5