请按照如下方式生成视频，用于给初中生讲解题目： 1.读题目并绘制图象 2.以上传图片中标准解析为参考，来讲解题目 3.总结题目的考查知识点、难度、易错点---**Extraction Content:** **Question Stem:** 如图，在等边△ABC中，D为AC中点，点P、Q分别为AB、AD上的点，BP = AQ = 4, QD = 2，在BD上有一点E，则PB+QE的最小值为 **Geometric Diagram Description:** Type: Geometric figure (triangle with points and lines). Main Elements: - Triangle ABC. - Point D is on side AC. - Point P is on side AB. - Point Q is on segment AD. - Segment BD is drawn. Point E is on segment BD. - Segments PQ, QE, and PE are drawn. - A dashed line segment QQ' is shown, crossing BD, suggesting reflection. Point Q' is labeled. In the diagram within the solution, Q' is shown on segment CD. - A dashed line segment PQ' is drawn, which intersects BD at point E. - A dashed line segment Q'E is drawn. - Vertices and points are labeled A, B, C, D, P, Q, E, and Q'. **Other Relevant Text (Standard Solution/Analysis):** 标准解析 解：如图，∵△ABC是等边三角形， ∴BA=BC, ∠A = 60° ∵D为AC中点， ∴BD⊥AC, 作点Q关于BD的对称点Q′，连接PQ′交BD于E，连接QE，此时PE+QE的最小值最小PE+QE = PE+EQ′=PQ′. ∴BP = AQ = 4, QD = 2, ∴AD = DC = AQ + QD = 6, QD = DQ′ = 2 ∴CQ′ = CD - DQ′ = 4 = BP, ∴AP = AQ′ = 8, ∴∠A = 60°, ∴△APQ′是等边三角形, ∴PQ′ = PA = 8, ∴PE + QE的最小值最小为8.