∵bcosB+ccosC=acosA,由正弦定理得:sinBcosB+sinCcosC=sinAcosA,即sin2B+sin2C=2sinAcosA,∴2sin(B+C)cos(B-C)=2sinAcosA.∵A+B+C=π,∴sin(B+C)=sinA.而sinA≠0,∴cos(B-C)=cosA,即cos(B-C)+cos(B+C)=0,∴2cosBcosC=0.∵0<B<π,0<C<π,∴B=90° 或C=90°,即△ABC是直角三角形.