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To solve the problem, we start by setting up coordinates for square (ABCD) with (A(0,0)), (B(1,0)), (C(1,1)), (D(0,1)). Let (E(1,e)) (on (BC)) and (F(f,1)) (on (CD)).

Step 1: Relate area of (\triangle AEF) to (e) and (f)

The area of (\triangle AEF) using cross product of vectors (\overrightarrow{AE}=(1,e)) and (\overrightarrow{AF}(f,1)) is:
[ \text{Area} = \frac{1}{2}\left|1 \cdot 1 - e \cdot f\right| = \frac{1}{2}\left|1 - ef\right| ]

Given the area is (\frac{1}{2}):
[ \left|1 - ef\right| = 1 \implies ef=0 \quad (\text{since } ef \leq 1)]

Step 2: Determine possible positions of (E) and (F)

(ef=0) implies either (e=0) ( (E=B) ) or (f=0) ( (F=D) ):

• If (E=B), then (F) must be (C) (to get area (\frac{1}{2})), so (\overrightarrow{AF}=(1,1)).
• If (F=D), then (E) must be (C), so (\overrightarrow{AE}=(1,1)).

Step 3: Calculate angle (\angle EAF)

For (\overrightarrow{AB}(1,0)) and (\overrightarrow{AC}(1,1)), the angle is (\tan^{-1}\left(\frac{1}{1}\right)=45^\circ).

This is the minimum possible angle (no smaller angle exists as other positions are invalid).

Answer: (\boxed{45^\circ}) or (\boxed{\frac{\pi}{4}}) (but typically (\boxed{45}) in competition settings).

(\boxed{45})

(\boxed{45})

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