(2) Make EF∨BC, prove the equilateral triangle AEF, and then prove △ DBE △ EFC to get the answer;
(3) There are two situations: one is that E is on the extension line of AB, the other is that E is on the extension line of BA, and D is on the extension line of BC, and CD= 1 can be obtained.
(1) The answer is: =.
(2) It is proved that in equilateral △ABC, ∠ ABC = ∠ ACB = ∠ BAC = 60, AB=BC=AC,
∫EF∨BC，
∴∠AEF=∠AFE=60 =∠BAC，
∴AE=AF=EF，
∴AB-AE=AC-AF，
That is, BE=CF,
∠∠ABC =∠EDB+∠BED = 60，
∠ACB =∠ ECB +∠FCE=60,
ED = EC，
∴∠EDB=∠ECB，
∠∠EBC =∠EDB+∠ Bede, ∠ ACB = ∠∠ European Central Bank +∠FCE,
∴∠BED=∠FCE，
∴△DBE≌△EFC，
∴DB=EF，
∴AE=BD.
(3)①∫AB = 1, AE=2, △ABC is an equilateral triangle, and B is the midpoint of AE.
∴AB=AC=BC= 1, easily available, △ACE is Rt△,
∴∠ACE=90，
∴∠ D =∠ ECB = 30, ∠ DBE =∠ ABC = 60, that is, △DEB is a right triangle.
∴ BD = 2 (the opposite side of 30 is equal to half of the hypotenuse), that is, CD = 1+2 = 3.
Another method: ∫EF∨CD
∴∠EFC=∠EBD= 180 -60
EC = ED
∴∠D=∠ECD，
Deb = ECF =60-equivalent cyclic density = 60-equivalent cyclic density
∴△EFC≌△EDB
∴EF=BD
∫∠A =∠ AEF again.
∴AE=2
∫BC = 1
∴CD=3
②∫AE = 2，BA=BC= 1，
∴BE=3, if EF⊥CD passes through CD at point F, then at Rt△EFB, ∠ BEF = 90-60 = 30,
∴bf=[ 1/2]be=[ 1/2]×( 1+3)= 1.5，
∴cf=bf-bc= 1.5- 1=0.5，
And ED=EC, EF⊥CD,
∴DF=CF (three lines in one),
∴CD=2CF= 1.
The length of a CD is 1 or 3.
Comments:
The test site of this topic: the judgment and nature of equilateral triangle; The nature of parallel lines; Congruent triangles's judgment and nature.
Test center comments: This topic mainly examines the understanding and mastery of knowledge points such as the nature and judgment of congruent triangles, the interior angle and theorem of triangle, and the nature and judgment of equilateral triangle. Comprehensive use of these properties for reasoning is the key to solve this problem.