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(B) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of sides $BC$,$CA$,and $AB$ respectively. By the midpoint theorem,the triangle is divided into f

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$\frac{1}{2}$

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(B) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of sides $BC$,$CA$,and $AB$ respectively. By the midpoint theorem,the triangle is divided into four congruent triangles: $\Delta AFE$,$\Delta FBD$,$\Delta EDC$,and $\Delta DEF$.Therefore,$	ext{Area}(\Delta AFE) = 	ext{Area}(\Delta FBD) = 	ext{Area}(\Delta EDC) = 	ext{Area}(\Delta DEF) = \frac{1}{4} 	ext{Area}(\Delta ABC)$.The quadrilateral $BDEF$ is composed of $\Delta FBD$ and $\Delta DEF$.Thus,$	ext{Area}(BDEF) = 	ext{Area}(\Delta FBD) + 	ext{Area}(\Delta DEF) = \frac{1}{4} 	ext{Area}(\Delta ABC) + \frac{1}{4} 	ext{Area}(\Delta ABC) = \frac{1}{2} 	ext{Area}(\Delta ABC)$.

In $\Delta ABC$,$D$,$E$ and $F$ are the midpoints of $\overline{BC}$,$\overline{CA}$ and $\overline{AB}$ respectively. Then,$\text{Area}(BDEF) = \ldots \times \text{Area}(\Delta ABC)$.

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