已知抛物线y=1/4x²+1和点A（0，2），点B是抛物线上一点，BC⊥x轴于点C，求证：AB=BC

设B点坐标为(x,y)那么y=1/4x^2+1BC=y=1/4x^2+1AB=√[(x-0)^2+(y-2)^2]=√[x^2+(1/4x^2+1-2)^2]=√[x^2+(1/4x^2-1)^2]=√(x^2+1/16x^4-1/2x^2+1)=√(1/16x^4+1/2x^2+1)=√(1/4x^2+1)^2=1/4x^2+1=BC