请按照图片中的解析思路，生成讲解视频，并总结题目考查知识点、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PE + QE的最小值为 **Chart/Diagram Description (associated with Question Stem):** * Type: Geometric figure (triangle). * Main Elements: * Triangle ABC (appears equilateral based on the problem statement). * Point D is the midpoint of AC. * Point P is on side AB. * Point Q is on segment AD. * Point E is on segment BD. * Segments PQ, PE, and QE are drawn. * Labels: A, B, C, D, P, Q, E. * Lines: Straight lines forming triangle ABC, and segments BD, PQ, PE, QE. * Shapes: Triangle ABC, various internal triangles (e.g., ABD, BCD, APQ, PEQ, QED, PEB, QEB). **Standard Solution:** 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PE + QE = PE + EQ' = PQ'. ∵BP = AQ = 4, QD = 2, ∴AD = DC = AQ + QD = 6, QD = DQ' = 2 ∴CQ' = CD - DQ' = 6 - 2 = 4 = BP, ∴AP = AB - BP. Since △ABC is equilateral, AB = AC = AD + DC = 6 + 6 = 12. So AP = 12 - 4 = 8. Also, from reflection, AQ' = AQ = 4. *Correction based on diagram and solution steps:* The solution states AP = AQ' = 8 later, implying AQ' might be 8, which contradicts AQ=4 and Q' being reflection of Q across BD. Let's re-examine the steps. The statement "∴AP = AQ' = 8" seems to be part of establishing △APQ' is equilateral. If AP=8 and ∠A=60°, then if △APQ' is equilateral, AQ' must also be 8. Let's follow the solution's derivation: AP = AB - BP = AB - 4. AQ = 4, QD = 2, so AD = 6. Since D is midpoint of AC, AC = 12. Since △ABC is equilateral, AB = BC = AC = 12. So AP = 12 - 4 = 8. Q is on AD, AQ = 4, QD = 2. Q' is the reflection of Q across BD. Since BD is the altitude from B to AC in equilateral △ABC, BD is also the angle bisector of ∠ABC and the median to AC. Reflection across the median BD in an equilateral triangle maps points on AC to points on AB. However, Q is on AD. Let's look at the diagram in the solution. Q is on AD. Q' is the reflection of Q across BD. BD is the line of symmetry for △ABC w.r.t the vertex B. Reflection across BD maps A to C and C to A. It maps points on AB to points on CB and points on CB to points on AB. Q is on AD. D is on AC. A is a vertex. Reflection of Q across BD would map Q to a point Q' on the line segment CD if Q was on AD and A was reflected to C. Let's verify Q' location from the diagram. Q' is shown in the diagram such that it seems to lie on CD extended, or somewhere near C. The solution states CQ' = CD - DQ'. This strongly suggests Q' is on CD. If Q is on AD and reflects to Q' on CD, then the line of reflection BD must pass through D, which it does. Also, BD must be the perpendicular bisector of QQ'. Since BD is the median and altitude in equilateral △ABC, BD ⊥ AC. D is the midpoint of AC. The reflection of Q across BD is Q'. Let's re-read the solution steps: "∵Q关于BD的对称点Q'," => QQ' ⊥ BD and the midpoint of QQ' lies on BD. Given Q is on AD, and D is the midpoint of AC. AD = DC = 6. AQ = 4, QD = 2. In △ABC, BD is the axis of symmetry through B. Reflection across BD maps △ABD to △CBD. Point A reflects to C. Point on AD would reflect to a point on CD. So, Q on AD reflects to Q' on CD. Since reflection preserves distances from the line of reflection and lengths, the distance of Q from BD is equal to the distance of Q' from BD. Also, DQ = DQ' = 2. Since Q is on AD and AQ=4, QD=2, AD=6. Q' is on CD and DQ'=2. CD=6. So CQ' = CD - DQ' = 6 - 2 = 4. BP = 4. Now, consider AP. AP = AB - BP. AB = AC = 12. AP = 12 - 4 = 8. AQ' = ? Triangle BDQ is reflected to triangle BDQ'. So BQ = BQ'. Also DQ = DQ' = 2. The solution states CQ' = CD - DQ' = 4 = BP. This is correct: CD=6, DQ'=2, so CQ'=4. BP=4. The solution then states AP = AQ' = 8. We calculated AP = 8. So AQ' must be 8. Let's check AQ'. In △ADQ, AD=6, AQ=4, QD=2. ∠ADQ = 180° (Q is on AD). This is incorrect. Q is *on* AD, so Q lies between A and D or coincides with A or D. Since AQ=4 and QD=2, Q is between A and D, and AD = AQ + QD = 4 + 2 = 6. D is midpoint of AC, so AD = DC = 6, AC = 12. △ABC is equilateral, so AB = BC = AC = 12. In △BDC, BD is the altitude, median, and angle bisector. When reflecting Q (on AD) across BD to get Q' (on CD), DQ = DQ' = 2. This is correct. The solution then states AQ' = 8. Let's verify this. In △ADC, BD ⊥ AC. D is midpoint of AC. In triangle ADQ, we have sides AD=6, AQ=4, QD=2. Q lies on AD, so A, Q, D are collinear. Consider triangle ABQ. AB=12, AQ=4, ∠A=60°. By Law of Cosines in △ABQ, BQ^2 = AB^2 + AQ^2 - 2(AB)(AQ)cos(60°) = 12^2 + 4^2 - 2(12)(4)(1/2) = 144 + 16 - 48 = 112. BQ = √112 = 4√7. Since Q' is reflection of Q across BD, BQ' = BQ = 4√7. Now let's look at AQ'. Q' is on CD, and DQ' = 2. A is a vertex. A, D, C are collinear. Q' is on CD, so A, D, Q', C are collinear in that order. AQ' = AD + DQ' = 6 + 2 = 8. This confirms AQ' = 8. So we have AP = 8 and AQ' = 8. The solution then considers △APQ'. AP = 8, AQ' = 8, ∠A = 60°. Since AP = AQ' and ∠A = 60°, △APQ' is an isosceles triangle with apex angle 60°. Therefore, △APQ' is an equilateral triangle. So, PQ' = AP = AQ' = 8. The minimum value of PE + QE is PQ'. Thus, the minimum value is 8. **Chart/Diagram Description (associated with Standard Solution):** * Type: Geometric figure (triangle) showing reflection. * Main Elements: * Triangle ABC (equilateral). * Point D is the midpoint of AC. * Point P is on side AB. * Point Q is on segment AD. * Segment BD is drawn (dashed in this diagram). * Point Q' is the reflection of Q across BD. Q' appears to be on CD. * Segment PQ' is drawn, intersecting BD at E. * Segment QE is drawn. * Segment QQ' is drawn (dashed). * Labels: A, B, C, D, P, Q, E, Q'. * Lines: Straight lines forming triangle ABC, and segments AD, CD, BD (dashed), AB, BC, PQ', QE, QQ' (dashed). * Shapes: Triangle ABC, various internal triangles. * Annotations: The point E is explicitly shown as the intersection of PQ' and BD. Q' is marked as the reflection of Q across BD. **Other Relevant Text:** * Section Titles: 题干 (Question Stem), 标准解析 (Standard Solution). * Solution steps involve geometric reasoning and calculations. No options (A, B, C, D) were present in the image.