$\ 证明:(1)∵∠A=∠A,∠ACD=∠B,∴△ACD∽△ABC$
$∴\frac{AD}{AC}=\frac {AC}{AB},∴ AC^{2}=AD×AB$
$(2)设AD=m(m＞0)$
$∵D为AB的中点，∴ AD=BD=m,AB=2m.由(1)得△ACD∽△ABC,∴\frac{CD}{BC}=\frac{AD}{AC}=\frac{AC}{AB}$
$∴AC^{2}=AD×AB=m×2m=2m^{2},∴ AC=\sqrt{2} m,∴ \frac{CD}{BC}=\frac{AC}{AB}=\frac{\sqrt{2}m}{2m}=\frac{\sqrt{2}}{2}$
$∵BC=4,∴CD=\frac{\sqrt{2}}{2}BC=\frac{\sqrt{2}}{2}×4=2\sqrt{2},∴CD的长是2\sqrt{2}$
$(3)过点B作BF⊥DC，交DC的延长线于点F，则∠F= 90°$
$∵E为CD的中点，∴CE=DE$
$设CE=DE=n(n＞0)$
$∵∠CDB=∠CBD=30°, ∴ CB =CD=2n,∠BCF=∠CDB+∠CBD=60°$
$∴ ∠FBC=90°-∠BCF=30°$
$∴CF= \frac{1}{2}CB =n$
$∴EF =CE +CF = 2n, BF =\sqrt{CB^{2}-CF^{2}}=\sqrt{3}n$
$∴ BD=2BF=2\sqrt{3}n,BE=\sqrt{EF^{2}+BF^{2}}= \sqrt{7}n$
$过点C作CH//EB,交AB 的延长线于点H,∴∠H=∠EBD,△HDC∽△BDE$
$∴\frac{HC}{BE}=\frac{HD}{BD}=\frac{CD}{ED}=\frac{2n}{n}=2$
$∴ HC=2BE=2\sqrt{7}n,HD=2BD=4\sqrt{3}n$
$∵∠ACD=∠EBD,∠H=∠EBD,∴∠ACD=∠H$
$∵∠A =∠A, ∴ △ACD∽△AHC,∴\frac{AD}{AC}=\frac{AC}{AH}=\frac{CD}{HC}=\frac{\sqrt{7}}{7}$
$∵AC=2\sqrt{7},∴AD=\frac{\sqrt{7}}{7}AC=2,AH=\sqrt{7}AC=\sqrt{7}×2\sqrt{7}=14$
$∴HD=AH-AD=14-2=12,∴ 4\sqrt{3}n=12，解得n=\sqrt{3}$
$∴ BE=\sqrt{7}×\sqrt{3}= \sqrt{21},∴BE的长是 \sqrt{21}$
