1. Let EF//AC pass through point E, so f is the midpoint of BC.

Because e is the midpoint of AB and triangle ABC is an equilateral triangle.

So CE is perpendicular to AB, AE=BE, so the angle ECB=30 degrees.

Yes, because DE=EC, angle D= angle BCE=30 degrees.

In the triangle DEC, the angle DEC =180-30-30 =120 degrees. So the angle DEB= angle DEC- angle BEC= 120-90=30 degrees.

So BD=BE, which means BD=AE.

According to the question, the triangle AEF is an equilateral triangle. So AE=EF, BE=CF

Because DE=EC and angle D= angle BCE.

And because EF//BC, angle FEC= angle BCE.

Triangle AEF and triangle ABC are equilateral triangles, so angle ABD= angle EFC= 120 degrees.

According to the angle, triangle BDE and triangle EFC are congruent, so BD=EF, so DB=AE.

3. Draw a picture according to the meaning of the question. AB= 1 AE=2, so b is the midpoint of AE. Triangle ABC is an equilateral triangle, so AB=AC=BC= 1. So the triangle ACE is a right triangle. The angle ACE is a right angle (according to the median line of the hypotenuse of a right triangle is equal to half of the hypotenuse), so the angle D= the angle ECB=30 degrees, and the angle DBE= the angle ABC=60 degrees, so the triangle DEB is a right triangle. So BD = 2 (the opposite side of 30 degrees is equal to half of the hypotenuse), so CD= 1+2=3.