A, B, C, D are any four points. If E and F ar | vedclass

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(D) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $E$ is the midpoint of $AC$, its positi

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$4 \vec{EF}$

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(D) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $E$ is the midpoint of $AC$, its position vector is $\vec{e} = \frac{\vec{a} + \vec{c}}{2}$, which implies $\vec{a} + \vec{c} = 2\vec{e}$.Since $F$ is the midpoint of $BD$, its position vector is $\vec{f} = \frac{\vec{b} + \vec{d}}{2}$, which implies $\vec{b} + \vec{d} = 2\vec{f}$.We need to evaluate the sum $\vec{S} = \vec{AB} + \vec{CB} + \vec{CD} + \vec{AD}$.Expressing these in terms of position vectors:$\vec{AB} = \vec{b} - \vec{a}$$\vec{CB} = \vec{b} - \vec{c}$$\vec{CD} = \vec{d} - \vec{c}$$\vec{AD} = \vec{d} - \vec{a}$Summing these up:$\vec{S} = (\vec{b} - \vec{a}) + (\vec{b} - \vec{c}) + (\vec{d} - \vec{c}) + (\vec{d} - \vec{a})$$\vec{S} = 2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c}$$\vec{S} = 2(\vec{b} + \vec{d}) - 2(\vec{a} + \vec{c})$Substituting the midpoint relations:$\vec{S} = 2(2\vec{f}) - 2(2\vec{e})$$\vec{S} = 4\vec{f} - 4\vec{e} = 4(\vec{f} - \vec{e}) = 4\vec{EF}$.

$A, B, C, D$ are any four points. If $E$ and $F$ are midpoints of $AC$ and $BD$ respectively, then $\vec{AB} + \vec{CB} + \vec{CD} + \vec{AD} =$

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