已知三角形ABC中BC=A,CA=b,AB=c,求角平分线AD的长

由AD是角平分线,得BD/CD=AB/AC=c/b.
又BD+CD=a,有BD=ac/(b+c),CD=ab/(b+c),
令AD=ta,则由斯特瓦尔特定理得
c^2*ab/(b+c)+b^2*ac/(b+c)-ta^2*a=a*ac/(b+c)*ab/(b+c),
∴ta^2=bc^2/(b+c)+b^2c*(b+c)-a^2bc/(b+c)^2=bc{1-[a/(b+c)]^2}.
定理自己查看下,我就不证明了.