Question

如图,AB是圆o的直径,CD是圆O的切线,切点为C.延长线AB交CD与E,连接AC,做∠DAC=∠ACD,做AF⊥ED与F,交圆O于G

如图,AB是⊙O的直径,CD是⊙O的切线,切点为C.延长线AB交CD与E,连接AC,做∠DAC=∠ACD,做AF⊥ED与F,交圆O于G
[1]求证:AD是圆O的切线(我已经证出来了)
[2]如果⊙O的半径=6,EC=8,求GF的长

Answer 1

∵∠BCE=∠CAE ∠E=∠E
∴△BCE相似△AEC
∴EB/EC=EC/EA即EB/8=8/EB+12
∴EB=4
∴EO=10
AE=16
∵△EOC相似△EAF
∴OC/AF=EO/EA
∴AF=906
∴EC/EF=EO/EA
∴EC/EF=EO/EA
∴8/8+CF=10/16 CF=4.8
∵∠CAF=∠GCF ∠AFC=∠GFC
∴△ACF相似△CGF
∴CF/AF=GF/CF
所以GF=2.4

Answer 2

EC^2 = EB*EA ==== EC^2 = EB(EB+AB) === 64 = EB^2 + 12EB === EB = 4

===AE = 16 ,OE = 10 === AF = AE*OC/OE = 48/5 , EC = 64/5 ====CF = 24/5

CF^2 = FG*FA ===......

追问

能告诉我用了哪些定理么,我自己研究下应该就看得懂你说的了...

追答

1,玄切线定理，2。三角形EOC，EAF相似，