若△ABC满足a(bcosB-ccosC)=(b^2-c^2)cosA,试判断△的形状

a(bCOSB-cCOSC)=(b^2-c^2)COSA,
而,
cosA=(b^2+c^c-a^2)/2bc,
cosB=(a^2+c^2-b^2)/2ac,
cosC=(a^2+b^2-c^2)/2ab,
把cosA,cosB,cosC代入a(bCOSB-cCOSC)=(b^2-c^2)COSA,中可得,
b^2*a^2-b^4-a^2*c^2+c^4=0,
a^2(b^2-c^2)-(b^4-c^4)=0,
(b^2-c^2)[a^2-(b^2+c^2)]=0,
b^2-c^2=0,a^2=b^2+c^2,
b=c,a^2=b^2+c^2,
即,三角形是等边直角三角形。