第1个回答 2010-05-12
证明:记AB=c,AC=b,BC=a.
由角平分线性质得:
AB/AC=BT/CT.
BT/TC=c/b
于是有:TC=ab/(b+c)=BS,BT=ac/(b+c)
由余弦定理,有:
AS^2=AB^2+BS^2-2*AB*BS*cosB
AT^2=AB^2+BT^2-2*AB*BT*cosB.两式相减,得:
AS^2-AT^2=a^2*(b-c)/(b+c)-2ac*COSB*(b-c)/(b+c)
=(b-c)[a^2/(b+c)-2ac*cosB/(b+c)]
=(b-c)^2*[(a^2-2ac*cosB)/(b^2-c^2)]=(b-c)^2=(AB-AC)²