本文将给出三角形中线长与角平分线长的一个有趣的几何不等式.定理在△ABC中,BC=a,CA=b,AB=c,ma是BC边上的中线长,wa是∠A的角平分线长,(1)当0A≤2π时,有wmaa≤(b+c)4b2c(b2+c2)(2)当2π≤Aπ时,有wmaa≥(b+c)4b2c(b2+c2)证明在△ABC中,由余弦定理知b2+c2-a2=2bccosA由三角? This paper gives an interesting geometric inequality of the length of the triangle centerline and the bisector of the angle. Theorem in △ABC, BC=a, CA=b, AB=c, ma is the midline length on the BC side, wa is ∠A The angle bisector is long. (1) When 0A≤2π, there is wmaa≤(b+c)4b2c(b2+c2)(2) When 2π≤Aπ, there is wmaa≥(b+c)4b2c(b2+ C2) prove that in △ABC, the cosine theorem knows b2+c2-a2=2bccosA from the triangle?