请生成动画视频解答这道题目---**Question 12:** **Question Stem:** 如图, 已知 AB=4, AC=2, 以 BC 为底边向上构造等腰 Rt△BCD, 连接 AD 并延长至点 P, 使 AD = PD, 则 PB 的取值范围为 **Mathematical Formulas/Equations in Question Stem:** AB = 4 AC = 2 AD = PD The range for PB is requested. **Other Relevant Text (Answer/Fill-in-the-blank):** 4-2√2 ≤ BP ≤ 4+2√2 **Chart/Diagram Description:** * **Type:** Geometric figure. * **Main Elements:** * **Points:** Labeled points A, B, C, D, P. * **Lines/Segments:** * Segments AB, AC, and BC form a triangle ABC. * Segments BD, CD, and BC form a triangle BCD. * Segment AD is drawn. * Segment AD is extended in a straight line to point P. * Segment BP is drawn, connecting point B to point P. * Implicit segment CP (connecting C to P) is formed by the points but not explicitly drawn. * **Shapes:** * Triangle ABC. * Triangle BCD. * Quadrilateral ABCD. * **Labels and Annotations:** * The length of segment AB is given as 4. * The length of segment AC is given as 2. * Triangle BCD is described as an isosceles right triangle ("等腰 Rt△BCD") constructed "upwards" from BC as its base. This implies that the angle at vertex D is the right angle (∠BDC = 90°), and the two legs BD and CD are equal in length (BD = CD). * Point D is the midpoint of segment AP, as indicated by the condition AD = PD. * **Relative Position:** Point D is positioned relative to BC to form the triangle BCD. Point P is collinear with A and D, with D situated between A and P. The overall arrangement suggests A, B, C form a base, and D, P extend outward/upward.