Question

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

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