过O作OE//BC交CD于点E,∵AD//BC,O为AB的中点∴E为CD的中点∵直角梯形ABCD，∠B=90°，AD//BC∴OE=(AD+BC)/2∵AD+BC=CD∴OE=CE=DE∴∠ODE=∠DOE,∠OCE=∠COE∵∠ODE+∠DOE+∠OCE+∠COE=180°∴∠DOC=∠DOE+∠COE=90°∴CD²=OD²+OC²∵OC=8㎝，OD=6㎝∴CD=10㎝
