∵CD=OD+OC,∴CD>OC。
∵AB is the diameter ⊙O, BC is the tangent ⊙O, ∴BC⊥OB, ∴ OC > BC. [The hypotenuse in RT delta delta is the largest]
From CD > OC, OC > BC, we get CD > BC, while CD = 2, BC = 3, ∴ 2 > 3.
This is of course wrong.
After correcting the data, OB (radius ⊙ O) can be obtained by the following methods:
∵ OC = OD-OD = CD-OB, and OB⊥BC, ∴ by Pythagoras theorem, there are: OC 2 = OB 2+BC 2.
∴(cd-ob)^2=ob^2+bc^2,∴cd^2-2cd×ob+ob^2=ob^2+bc^2，
∴2cd×ob=cd^2-bc^2,∴ob=(cd^2-bc^2)/(2cd)。
The second question:
∵AB is the diameter ∵ O, ∴∠ ADB = 90, ∴∠∠ EDB = 90.
From the proof process of the first question, there are: ∠ EBA = 90.
∴∠ EDB =∠ EBA = 90，∴△ebd∽△eab bed =∠aeb，∴∠ FBD = ∠ bad。
∵BD⊥ED、BF=EF,∴DF=BE/2=BF,∴∠FBD=∠FDB。
From ∠ FBD = ∠ bad, ∠ FBD = ∠ FDB, the tangent angle of ∠ FDB = ∠ bad, ∠ FDB is ∠ O is obtained.
∴DF is the tangent of⊙ O.
Note: In the following links, the answer to the first question is wrong and the relevant data is unreasonable.