如图∠abc=∠acb,be平分∠abc,cf平分∠acb;∠ebd=∠d,试猜想cf与de的关系!!!!!!!!!急...过程

如图∠abc=∠acb,be平分∠abc,cf平分∠acb;∠ebd=∠d,试猜想cf与de的关系

猜想:
①CF // DE
证：∵BE平分∠ABC, CF平分∠ACB
∴∠EBC = 1/2∠ABC
∠BCF = 1/2∠ACB

∵∠ABC = ∠ACB
∴∠EBC = ∠BCF

∵∠EBD(∠EBC) = ∠D
∴∠BCF = ∠D
∴CF // DE

②CF = DE
证：∵BE平分∠ABC, CF平分∠ACB
∴∠ABE = 1/2∠ABC
∠ACF = 1/2∠ACB

∵∠ABC = ∠ACB
∴∠ABE = ∠ACF, AB = AC

∵∠A = ∠A
∴△ACF ≌ △ABE
∴BE = CF

∵∠EBD = ∠D
∴BE = ED
∴CF = ED

 先根据∠abc=∠acb，be平分∠abc，cf平分∠acb，可求出∠ebd=∠bcf，再根据∠ebd=∠d可求出∠bcf=∠d，进而可求出cf∥de．   解：cf∥de．理由如下： ∵∠abc=∠acb，be平分∠abc，cf平分∠acb， ∴∠ebd=∠bcf， ∵∠ebd=∠d， ∴∠bcf=∠d， ∴cf∥de．   本题考查了角平分线的性质及平行线的判定定理，比较简单．