怎么做---Question Number: 15 Question Stem: 如图，在矩形 ABCD 中，AB=2, AD=4, 观察图中尺规作图的痕迹，则 DP 的长为 ______． English Translation: As shown in the figure, in rectangle ABCD, AB=2, AD=4. Observe the traces of compass and ruler construction in the figure, then the length of DP is ______. Options: (No options provided) Diagram Description: * Type: Geometric diagram. * Main Elements: * A rectangle ABCD with vertices labeled in counter-clockwise order starting from top left (A at top left, B at bottom left, C at bottom right, D at top right). * Line segments forming the sides of the rectangle: AB, BC, CD, DA. * A diagonal line segment AC is drawn. * A point P is labeled on the side CD. * Construction Marks: * An arc is drawn centered at vertex A, passing through vertex D. This arc intersects the diagonal AC at a point (not explicitly labeled, but implied by the construction). Let's denote this intersection point as E. The radius of this arc is AD. * An arc is drawn centered at vertex C. The radius of this arc appears to be the distance from C to the intersection point E on AC. This arc intersects the side CD at point P. Thus, the distance CP is equal to the distance CE. * Labels and Annotations: Vertices A, B, C, D, P are labeled. The side lengths are given in the question stem. The figure is labeled "第 15 题图" (Figure for Question 15) below the diagram. Other Relevant Text: 第 15 题图 (Figure for Question 15) Mathematical Formulas/Relationships derived from the problem and diagram: * Rectangle ABCD: AB || CD, AD || BC, all angles are 90 degrees. * Given lengths: AB = 2, AD = 4. * In a rectangle, opposite sides are equal: CD = AB = 2, BC = AD = 4. * By Pythagorean theorem in right triangle ADC: $AC^2 = AD^2 + CD^2 = 4^2 + 2^2 = 16 + 4 = 20$. So, $AC = \sqrt{20} = 2\sqrt{5}$. * Construction 1: Arc centered at A through D intersects AC at E. Thus, AE = AD = 4. * Point E is on AC, so AC = AE + EC. Therefore, EC = AC - AE = $2\sqrt{5} - 4$. * Construction 2: Arc centered at C with radius CE intersects CD at P. Thus, CP = CE = $2\sqrt{5} - 4$. * Point P is on the line segment CD. DP = CD - CP. * DP = $2 - (2\sqrt{5} - 4) = 2 - 2\sqrt{5} + 4 = 6 - 2\sqrt{5}$.