三角形ABC,AP为∠A平分线
AP=t,AB=c,AC=b,BC=a,PC=X,BP=A-X=y
b/c=x/y=x/(a-x)
x=ab(b+c),y=ac/(b+c)
cos∠APB=-cos∠APC
余弦定理:
(x^2+t^2-b^2)/(2tx)=-(y^2+t^2-c^2)/(2ty)
整理:
(x+y)t^2=(xc^2+yb^2)-xy(x+y),x+y=a
t^2
=(xc^2+yb^2)/a-xy
=bc(b+c)/(b+c)-xy
=bc-xy
即:
AP^2=AB*AC-BP*PC
所以:
三角形内角平分线的平方,等于两条邻边的乘积,减去对边被角平分线分成的两条线段的积