如圖,作OE⊥AC、OF⊥BD,分別連接OB、OM、OC.
則：OE²=OC²-CE², OF²=ME²=OM²-OE²=OM²-(OC²-CE²)=OM²+CE²-OC²,
    BF²=OB²-OF²=OB²-(OM²+CE²-OC²)=OB²+OC²-OM²-CE²=2(OB)²-OM²-CE².
由題意知：OB=2、 OM=√3 ,故：BF=√(5-CE²).
則：AC+BD=2CE+2BF=2(CE+BF)=2[CE+√(5-CE²)]
由不等式x+y≤√[2(x²+y²)]得：CE+√(5-CE²)≤√[2(CE²+5-CE²)=√10.
所以：AC+BD≤2√10,即AC+BD的最大值為2√10.