解:A—D—M—C—N—B
∵M是AC的中点
∴AM=AC/2=(AB-BC)/2
∵N是BD的中点
∴DN=BD/2=(AB-AD)/2
∴AN=AD+DN=AD+(AB-AD)/2=(AB+AD)/2
∴MN=AN-AM=(AB+AD)/2-(AB-BC)/2=(AD+BC)/2=(AB-CD)/2
∵AB=a,CD=b,
∴MN=(a-b)/2 (厘米)
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