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Heping district sanmo mathematics
Solution: Connect be.

∵BC is the tangent of ∵ O ∴∠ ABC = 90.

∫ab is∫ ∫o∴∠aeb = 90°.

∴∠DBE+∠OBE=90,∠AEO+∠OEB=90

∵OB=OE,∴∠OBE=∠OEB∴∠DBE=∠AEO?

∵∠AEO=∠CED∴∠CED=∠CBE,

∵∠C=∠C,∴△CED∽△CBE,

∴CE2=CD? CB?

∵ob= 1,bc=2,∴oc=5,∴ce=oc-oe=5- 1

(5- 1)2=2CD,∴CD=3-5.

So the answer is: 3-5.