请帮忙生成该题的解题思路视频，要求有完整的解题过程和正确的解题思路---**General Problem Description:**
在△ABC中, ∠BAC = 90°, AB = AC = 2√2, D 为 BC 上任意一点, E 为 AC 上任意一点.
**Figure 1 Description:**
* Type: Geometric figure (Triangle).
* Elements:
* Triangle ABC with right angle at A. Vertices B, A, C labeled.
* Point D is on line segment BC.
* Point E is on line segment AC.
* Line segment DE is drawn.
* Labels: A, B, C, D, E. "图1" is below the figure.
**Question (1):**
(1) 如图 1, 连接 DE, 若 ∠CDE = 60°, AC = 4AE, 求 DE 的长.
**Figure 2 Description:**
* Type: Geometric figure (Triangle with internal lines).
* Elements:
* Triangle ABC (likely the same base triangle as in Figure 1, but not explicitly stated). Vertices B, A, C labeled.
* Point D is on line segment BC.
* Line segment AD is drawn.
* Point F is on line segment AD.
* Line segment EF is drawn and extended to intersect line AB at point M.
* Line segment EG is drawn, appearing to be related to EF via rotation around E.
* Line segment AG is drawn.
* Point N is on line segment AC.
* Line segment GN is drawn.
* Various intersections form internal points/lines (e.g., intersection of AG and FN). Point G is outside triangle ABC.
* Labels: A, B, C, D, E, F, G, M, N. "图2" is below the figure.
**Question (2):**
(2) 如图 2, 若点 D 为 BC 中点, 连接 AD, 点 F 为 AD 上任意一点, 连接 EF 并延长交 AB 于点 M, 将线段 EF 绕点 E 顺时针旋转 90° 得到线段 EG, 连接 AG. 点 N 在 AC 上, ∠AGN = ∠AEG 且 AM + AF = √2AE, 求证: GN = MF.
**Figure 3 Description:**
* Type: Geometric figure (Triangle with internal lines).
* Elements:
* Triangle ABC (likely the same base triangle as in Figure 1, but not explicitly stated). Vertices B, A, C labeled.
* Point D is on line segment BC.
* Line segment AD is drawn.
* Point F is on line segment AD.
* Line segment EF is drawn.
* Line segment BF is drawn.
* Line segment EF is rotated around E to get EG.
* Line segment AG is drawn.
* Point H is on line AB.
* Line segment FH is drawn.
* Point G is outside triangle ABC.
* Labels: A, B, C, D, E, F, G, H. "图3" is below the figure.
**Question (3):**
(3) 如图 3, 点 D 为 BC 中点, 连接 AD, 点 F 为 AD 的中点, 连接 EF、BF, 将线段 EF 绕点 E 顺时针旋转 90° 得到线段 EG, 连接 AG, H 为直线 AB 上一动点, 连接 FH, 将 △BFH 沿 FH 翻折至 △ABC 所在平面内, 得到 △B'FH, 连接 B'G, 直接写出线段 B'G 的长度的最大值.