如图,在⊙O中,CD是直径,弦AB交CD于点M，且C是弧ACB的中点，ME⊥AC于点E，AC=5，且CE∶EA=3∶2

图画得不太标准。。。                               /    这边是不用相似做的

不知道你学没学相似三角形                        /∵AC=5  CE:AE=3:2 E在AC上

∵AC=5  CE:AE=3:2 E在AC上                    /∴CE=3 AE=2

∴CE=3 AE=2                                             /∵CD是直径  C是弧ACB的中点

∵CD是直径  C是弧ACB的中点                  我∴CD⊥AB CD平分AB 

∴CD⊥AB CD平分AB                                是  设AM=x  则CM=根号下(25-x²)

∵∠EAB+∠=90°=∠EAB+∠MCA              分  EM²=CM²-CE²=AM²-AE²

∴∠AME=∠MCA                                       割        25-x²-9=x²-4

又∵ME⊥AC于点E                                     线           X=√10

∴∠AEM=∠MEC                                        /              AM=√10 

∴△AEM∽△MEC                                       /         AB=2AM=2√10

AE:EM=EM:CE                                             /

EM=√6  AM=√10                                          /

AB=2AM=2√10                                            /