To solve the problem, we start by assigning coordinates to the square and analyzing the sum of distances from point (P) (on line (EF)) to the vertices (A, B, C, D).

Step 1: Coordinate Setup

Let the square have vertices:
(A(0,0)), (B(1,0)), (C(1,1)), (D(0,1)).

• (E) (midpoint of (AD)): ((0, 0.5))
• (F) (midpoint of (BC)): ((1, 0.5))

Line (EF) is (y=0.5) ((x \in [0,1])), so any (P) on (EF) is ((t, 0.5)) for (t \in [0,1]).

Step 2: Sum of Distances

The sum (PA + PB + PC + PD) simplifies because:

• (PA = PD) (symmetry)
• (PB = PC) (symmetry)

Thus:
(PA + PB + PC + PD = 2(PA + PB))

(PA = \sqrt{t^2 + (0.5)^2}), (PB = \sqrt{(1-t)^2 + (0.5)^2}).

Step 3: Minimize the Sum

The sum (PA + PB) is minimized when (t=0.5) (by symmetry and calculus). At (t=0.5), (P) is the center ((0.5, 0.5)):

(PA = \sqrt{(0.5)^2 + (0.5)^2} = \frac{\sqrt{2}}{2})

(PA + PB = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2})

Total sum: (2 \times \sqrt{2} = 2\sqrt{2}).

Answer: (\boxed{2\sqrt{2}})

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