为简化计算,把M(1,√2）绕O旋转至N(0,√3）,
设AC:kx-y+√3=0,则
BD:x+k(y-√3）=0,
O到AC的距离d1=（√3）/√（k^2+1),
O到BD的距离d2=|k√3|/√（k^2+1),
d1^2+d2^2=3,
(d1d2)^2<=(3/2)^2,
AC=2√（4-d1^2),
BD=2√（4-d2^2),
设w=AC+BD,
则(w/2)^2=8-（d1^2+d2^2)+2√[(4-d1^2)(4-d2^2)]
=5+2√[16-4(d1^2+d2^2)+(d1d2)^2]
<=5+2√[16-4*3+9/4]=5+5=10,
∴w/2<=√10,
w<=2√10,当d1=d2=√(3/2)时取等号,
∴AC+BD的最大值是2√10.