根据图片中的题目、答案与解析制作一个讲解视频，要求几何图形能动起来，知识点符合七年级数学---Here is the complete and accurate extraction of the content from the image, presented in a structured plain text format. **Question Stem:** 如图，在长方形 ABCD 中，AD // BC, E 为边 BC 上一点，将长方形沿 AE 折叠，使点 B 落在点 F 处，EG 平分 ∠CEF, 交 CD 于点 G, 过点 G 作 HG ⊥ EG, 交 AD 于点 H. (1)试说明：HG // AE. (2)若 ∠EAF=20°, 求 ∠DHG 的度数. **Chart/Diagram Description:** * **Type:** Geometric Figure (Rectangle with internal lines and folds). * **Main Elements:** * **Rectangle ABCD:** Vertices are labeled A, B, C, D in counter-clockwise order. Sides AB, BC, CD, DA form the rectangle. Note that the lines AB and DC are depicted as dashed lines in the diagram, while AD and BC are solid. However, the problem statement refers to ABCD as a rectangle, implying all sides are solid lines. This might be a visual convention to emphasize the interior elements. The line HG is parallel to AE. * **Point E:** Located on side BC. * **Line Segment AE:** This is the fold line. * **Point F:** The location where point B lands after folding along AE. * **Line Segment EF:** Connects E to F. * **Line Segment EG:** Extends from E to G, bisecting ∠CEF. * **Point G:** Located on side CD. * **Line Segment HG:** Drawn from G, intersecting AD at H. * **Right Angle:** A square symbol indicates that ∠HGE is a right angle (HG ⊥ EG). * **Point H:** Located on side AD. * **Annotation:** 第 5 题图 (Figure for Question 5) **Solution:** 解: (1) 由折叠，得 ∠AEB = ∠AEF. ∵ EG 平分 ∠CEF, ∴ ∠FEG = ∠CEG. ∵ ∠AEB + ∠AEF + ∠FEG + ∠CEG = 180°, ∴ ∠AEG = ∠AEF + ∠FEG = 1/2 × 180° = 90°, ∴ AE ⊥ EG. 又 ∵ HG ⊥ EG, ∴ HG // AE. (2) ∵ ∠EAF = 20°, ∴ ∠BAE = 20°, ∴ ∠DAE = 90° - ∠BAE = 70°. 又 ∵ HG // AE, ∴ ∠DHG = ∠DAE = 70°.