由D在线段BC上,可设AD = tAB+(1-t)AC,其中0 < t < 1.
由三角形内角平分线性质定理,|BD|:|DC| = c/b.
BD = AD-AB = (1-t)BC,DC = AC-AD = tBC.
于是(1-t)/t = c/b,有t = b/(b+c),AD = (bAB+cAC)/(b+c).
AD² = (b²AB²+c²AC²+2bcAB·AC)/(b+c)² = (2b²c²+2bcAB·AC)/(b+c)².
而BC² = (AC-AB)² = AC²+AB²-2AB·AC,即2AB·AC = b²+c²-a².
代入得AD² = bc(2bc+b²+c²-a²)/(b+c)² = bc((b+c)²-a²)/(b+c)².
另外(1)也可以由夹角相等得到等式bAB·AD = cAC·AD.
代入AD = tAB+(1-t)AC可解得t = b/(b+c) (可约去一个非零因子bc-AB·AC).