在三角形ABC中,若b/（cosA/2)=b/(cosB/2)=c/(cosC/2),则三角形ABC的形状是?

用正弦定理：sinA/a = sinB/b = sinC/c因为：a/(cosA/2) = b/(cosB/2) = c/(cosc/2)所以：sinA/(cosA/2) = sinB/(cosB/2) = sinC/(cosc/2)再利用倍角公式：2*sin(A/2)*cos(A/2)/(cosA/2) = 2*sin(B/2)*cos(B/2)/(cosB/2) = 2*sin(C/2)*cos(C/2)/(cosc/2)所以：sin(A/2) = sin(B/2) = sin(C/2)由于 0 < A,B,C < pi/2,sin函数在（0,pi/2）上单调所以：A = B = C所以是等边三角形注：pi是圆周率