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                                            /