设M(a,1/a),则D(a,-a+m),C( m-1/a ,1/a); A(0,m),B(m,0);
AD=√[a^2+(-a+m-m)^2]=a√2;
BC=√[m-1/a-m)^2+(0-1/a)^2]=√2/a
所以AD×BC=a√2×(√2/a)=2
设M(a,1/a),则D(a,-a+m),C( m-1/a ,1/a); A(0,m),B(m,0);
AD=√[a^2+(-a+m-m)^2]=a√2;
BC=√[m-1/a-m)^2+(0-1/a)^2]=√2/a
所以AD×BC=a√2×(√2/a)=2