As shown in Figure 9ab, the questions in Dalian Middle School are the diameter of circle O, the tangent between cd and circle O, and the extension line between cda and circle O perpendicular to abdo a

As shown in Figure 9ab, the questions in Dalian Middle School are the diameter of circle O, the tangent between cd and circle O, and the extension line between cda and circle O perpendicular to abdo and do. Analysis:

(1) If AB is the diameter of ⊙O and AB is ⊥ da, the tangent with AD ⊙O and DC ⊙O can be obtained, and the answer can be obtained according to the tangent length theorem;

(2) Connect BF, CE and AC, find DC=DA=4 by tangent length theorem, find the lengths of DO=5, CM and AM, find the length of BC by pythagorean theorem, and find CG/GF=BC/EF=3/5 according to △BGC∽△FGE, then CG = 3/8cf; Find the length of CF by Pythagorean theorem, and then you can find the length of CG.

Answer:

(1) Proof:

∵AB is the diameter ⊙ O, AB⊥DA,

∴AD is the tangent of⊙ o,

∵DC⊙o tangent,

∴DA=DC.

(2) Solution: Connect BF, CE, AC,

/& gt； From the tangent length theorem: DC=DA=4, DO⊥AC

∴DO bisects AC, at Rt△DAO, AO=3, AD=4, which is obtained by Pythagorean theorem: DO=5.

From the triangle area formula: 1/2DA? AO= 1/2DO？ AM，

Then AM= 12/5, similarly CM=AM= 12/5,

AC=24/5。

∫AB is the diameter,

∴∠ACB=90，

According to Pythagorean theorem, BC = √ [6 2-(24/5) 2] =18/5.

∠∠GCB =∠GEF, ∠GFE=∠GBC, (Angle theorem of circle)

∴△BGC∽△FGE，

∴cg/eg=bc/ef= 18/(5/6)=3/5

In Rt△OMC, CM= 12/5, OC=3, which is obtained by Pythagorean theorem: OM=9/5.

In Rt△EMC, CM= 12/5, ME=OE-OM=3-9/5=6/5, which is obtained by Pythagorean theorem: CE=(6/5)√5,

In Rt△CEF, EF=6, CE=6/5√5, which is given by Pythagorean theorem: cf = (12/5) √ 5.

∫CF = CG+GF，CG/EG=3/5，

∴cg=3/8cf=3/8×[( 12/5)√5]=(9/ 10)√5].