如图,△ABC中,AB>AC,AT是∠BAC的平分线,在BC上有一点S,是BS=TS,求证:AS²-AT²=(AB-AC)²打错了,是BS=TC
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![如图,△ABC中,AB>AC,AT是∠BAC的平分线,在BC上有一点S,是BS=TS,求证:AS²-AT²=(AB-AC)²打错了,是BS=TC](/uploads/image/z/6931109-29-9.jpg?t=%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E4%B8%AD%2CAB%EF%BC%9EAC%2CAT%E6%98%AF%E2%88%A0BAC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E5%9C%A8BC%E4%B8%8A%E6%9C%89%E4%B8%80%E7%82%B9S%2C%E6%98%AFBS%3DTS%2C%E6%B1%82%E8%AF%81%EF%BC%9AAS%26%23178%3B-AT%26%23178%3B%3D%EF%BC%88AB-AC%EF%BC%89%26%23178%3B%E6%89%93%E9%94%99%E4%BA%86%EF%BC%8C%E6%98%AFBS%3DTC)
如图,△ABC中,AB>AC,AT是∠BAC的平分线,在BC上有一点S,是BS=TS,求证:AS²-AT²=(AB-AC)²打错了,是BS=TC
如图,△ABC中,AB>AC,AT是∠BAC的平分线,在BC上有一点S,是BS=TS,求证:AS²-AT²=(AB-AC)²
打错了,是BS=TC
如图,△ABC中,AB>AC,AT是∠BAC的平分线,在BC上有一点S,是BS=TS,求证:AS²-AT²=(AB-AC)²打错了,是BS=TC
记AB=c,AC=b,BC=a
则由角平分线的性质:
BT:TC=c:b
BT+TC=a
于是有:
TC=ab/(b+c)=BS,
BT=ac/(b+c)
SC=a-BS=ac/(b+c)
过点A做BC垂直线交BC于H,
HC=(a^2+b^2-c^2)/(2a)
勾股定理可得
AS^2-AT^2=(AH^2+SH^2)-(AH^2+TH^2)
=SH^2-TH^2
=(SC-HC)^2-(TC-HC)^2
=SC^2-TC^2-2HC(SC-TC)
=[ac/(b+c)]^2-[ab/(b+c)]^2-2[(a^2+b^2-c^2)/(2a)][(ac-ab)/(b+c)]
=a^2(c^2-b^2)/(b+c)^2-(a^2+b^2-c^2)(c^2-b^2)/(b+c)^2
=(c^2-b^2)^2/(b+c)^2
=(c-b)^2
=(AB-AC)^2
http://zhidao.baidu.com/link?url=B_SheywTJwT6gadnqczxVCOV7wznbA-IUOwNS_0fuLmaX4GKTAz7Eq6RDTPg5nG4NSC4EngifEuBCmcFkJiraq
这有个类似的,你参考下吧