设两方程为y=ax+b y=k/x则C(0,b) D(-b/a,0)设A、B的坐标为(x1,y1) (x2,y2)y=ax+b代入y=k/x得ax^2+bx-k=0知x1+x2=-b/a (1)要证AC=BD即√[x1^2+(y1-b)^2]=√[(x2+b/a)^2+y2^2]平方,并把y1=ax1+b y2=ax+b代入化简即要证[a(x1+x2)+b](ax2-ax1+b)=0 (2)将(1)代入(2)得知(2)成立因此证得AC=BD