几何作图，求解这道题---**Extraction Content:** **Question Stem:** 如图, 正方形ABCD边长为4, 在Rt△EFG组, ∠FEG=90°, EF=FG=2√2 D是斜边FG中点, 连接BE, P为BE中点, 连接PC, 求: PC的最大和最小值? **Other Relevant Text:** 求PC的最大最小值? @阿古数学 (Watermark) **Chart/Diagram Description:** * **Type:** Geometric diagram. * **Main Elements:** * Square ABCD with vertices labeled A, B, C, D. Side length is 4. The vertices are labeled as A top-left, B bottom-left, C bottom-right, D top-right. * A right triangle EFG with a right angle at E (∠FEG = 90°). The text states side lengths EF = FG = 2√2. (Note: In a Euclidean right triangle, the hypotenuse FG should be longer than the leg EF if E is the right angle, indicating a potential typo in the text. Assuming the intended triangle is an isosceles right triangle at E, where EF=EG=2√2, making FG=4. In this case, the text "EF=FG=2√2" is contradictory). * Point D: The text states that D (vertex of the square) is the midpoint of the hypotenuse FG of triangle EFG. If E is the right angle, FG is the hypotenuse. If D is the midpoint of FG, then the distance from D to E, F, and G are equal (DE = DF = DG = FG/2). Based on the likely intended triangle (EF=EG=2√2, ∠FEG=90°, FG=4), DE would be 4/2 = 2. So E lies on a circle centered at D with radius 2. * Point E: Shown inside or on the boundary of the square. * Point P: Midpoint of the line segment BE. * Line segments BE and PC are drawn. * **Labels and Annotations:** Points A, B, C, D, E, F, G, P are labeled. The watermark "@阿古数学" is present over the square. * **Relative Position and Direction:** The square ABCD is the main reference frame. The triangle EFG is positioned relative to the square such that vertex D of the square is the midpoint of the hypotenuse FG, and vertex E is inside or on the boundary of the square. P is on BE, and C is a vertex of the square.