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(C) Let $D, E, F$ be the mid-points of sides $AB, BC, CA$ respectively of $	riangle ABC$.By the Mid-point Theorem,$DE \parallel AC$ and $DE = \frac{1

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$\frac{1}{2} \operatorname{ar}(	riangle ABC)$

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(C) Let $D, E, F$ be the mid-points of sides $AB, BC, CA$ respectively of $	riangle ABC$.By the Mid-point Theorem,$DE \parallel AC$ and $DE = \frac{1}{2} AC = AF$.Similarly,$EF \parallel AB$ and $EF = \frac{1}{2} AB = AD$.Thus,$ADEF$ is a parallelogram.Since the line segment joining the mid-points of two sides of a triangle divides the triangle into four congruent triangles,each having an area equal to $\frac{1}{4}$ of the area of the original triangle.Therefore,$\operatorname{ar}(	riangle ADE) = \operatorname{ar}(	riangle BDE) = \operatorname{ar}(	riangle EFC) = \operatorname{ar}(	riangle AFE) = \frac{1}{4} \operatorname{ar}(	riangle ABC)$.The parallelogram $ADEF$ is composed of two triangles,$	riangle ADE$ and $	riangle AFE$.Thus,$\operatorname{ar}(ADEF) = \operatorname{ar}(	riangle ADE) + \operatorname{ar}(	riangle AFE) = \frac{1}{4} \operatorname{ar}(	riangle ABC) + \frac{1}{4} \operatorname{ar}(	riangle ABC) = \frac{1}{2} \operatorname{ar}(	riangle ABC)$.

The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

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