Question

如图,⊙O中,弦AB、CD相交于AB的中点E,连接AD并延长至点F,使DF=AD,连接BC、BF．

Answer 1

∠CEB=∠AED,

AE=EB,AD=DF,

AB=2AE,AF=2AD,

AE:AB=AD:AF,∠EAD=∠DAF,

△EAD∽△BAF,

∠AED=∠ABF=∠CED,

∠BCE=∠BAF,(同弧所对圆周角相等),

故∠CBE=∠AFB,

因此△CBE∽△AFB(AAA)；

BE/FB=CB/AF=CB/(2AD),

CB/AD=2BE/FB=2*5/8=5/4.