请解答图片中的题目---**Question 2**
**Question Stem:**
如图, 等边三角形 ABC 中, D 为 AB 边上一点 (点 D 不与点 A, B 重合), 连接 CD, 将 CD 平移到 BE (其中点 B 和 C 对应), 连接 AE. 将 △BCD 绕着点 B 逆时针旋转至 △BAF, 延长 AF 交 BE 于点 G.
**Sub-questions:**
(1) 连接 DF, 求证: △BDF 是等边三角形;
(2) 求证: D, F, E 三点共线;
(3) 当 BG=2EG 时, 求 tan∠AEB 的值.
**Diagram Description:**
* **Type:** Geometric figure.
* **Main Elements:**
* **Points:** A, B, C, D, E, F, G.
* **Shapes:** Triangle ABC (labeled as equilateral), Triangle BCD, Triangle BE (formed by translation), Triangle BAF (formed by rotation), Triangle BDF (formed by connecting DF), Triangle ABE, Triangle AGE, Triangle ABG.
* **Lines:**
* Sides of triangle ABC: AB, BC, CA.
* Line segment CD.
* Line segment BE (translation of CD).
* Line segment AE.
* Line segment DF (connecting D and F).
* Line segment AF is extended to intersect BE at G. Thus, points A, F, G are collinear.
* Points E, G, B are collinear as G is on BE.
* **Relative Position and Direction:**
* Triangle ABC is equilateral.
* Point D is on side AB, distinct from A and B.
* Line segment BE is a translation of CD, such that B corresponds to C and E corresponds to D. This implies vector $\vec{BE} = \vec{CD}$. Since B corresponds to C, it means B is the image of C under translation, and E is the image of D. However, the description says "点 B 和 C 对应", which typically means B in BE corresponds to C in CD, and E in BE corresponds to D in CD. This means the translation vector is $\vec{CB}$. So $\vec{BE} = \vec{CB}$. This implies B is the midpoint of CE, or C, B, E are collinear with CB = BE. This interpretation conflicts with the diagram. A more common interpretation of "将 CD 平移到 BE (其中点 B 和 C 对应)" in geometry problems of this type is that segment CD is translated such that C moves to B, and D moves to E. This would mean $\vec{BE} = \vec{CD}$. Let's assume $\vec{BE} = \vec{CD}$.
* Triangle BCD is rotated around point B counterclockwise to triangle BAF. This means B is the center of rotation, BC rotates to BA, and BD rotates to BF. The angle of rotation is ∠CBA, which is 60° since △ABC is equilateral. Therefore, BC = BA, BD = BF, and ∠CBD = ∠ABF, and ∠CBC = ∠ABA = 60°. Wait, ∠CBA = 60°. So rotation angle is 60°. This means BD rotates to BF and ∠DBF = 60°. Also BC rotates to BA and ∠CBA = 60°. BC=BA is given by equilateral triangle. So under rotation around B by 60° counterclockwise, C maps to A, and D maps to F. Thus, B-C-D maps to B-A-F. This means BC=BA, BD=BF, and CD=AF. Also ∠CBD rotates to ∠ABF, so ∠CBD = ∠ABF. The angle of rotation is ∠CBA = 60°. Therefore, ∠DBF = 60°.
* Point G is the intersection of the extension of AF and the line segment BE.
* **Labels and Annotations:** Points A, B, C, D, E, F, G are labeled.
*(Note: The description of the translation "将 CD 平移到 BE (其中点 B 和 C 对应)" is slightly ambiguous in the original text, but the diagram and subsequent rotation information strongly suggest the interpretation where C translates to B, and D translates to E, i.e., $\vec{BE} = \vec{CD}$. The rotation information "将 △BCD 绕着点 B 逆时针旋转至 △BAF" implies C maps to A and D maps to F under a rotation around B by 60 degrees counterclockwise, which is consistent with ABC being equilateral. This means BC=BA, BD=BF, CD=AF, and ∠CBD = ∠ABF, ∠DBF = 60°. The translation part BE=CD and BE is parallel to CD. Combining with BE=AF from rotation, we get BE=AF. And the line passing through D, F, E is suggested by question (2).)*