If D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC, write the value of $\vec{A D} + \vec{B E} + \vec{C F} .$

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Solution

Given: $D, E, F$ are the midpoints of the sides $B C, C A, A B$ respectively. Then, the position vectors of the midpoints $D, E, F$ are given by $\frac{\vec{b} + \vec{c}}{2}, \frac{\vec{c} + \vec{a}}{2}, \frac{\vec{a} + \vec{b}}{2}$
$Now, \vec{A D} + \vec{B E} + \vec{C F} = (\frac{\vec{b} + \vec{c}}{2}) - \vec{a} + (\frac{\vec{c} + \vec{a}}{2}) - \vec{b} + (\frac{\vec{a} + \vec{b}}{2}) - \vec{c} = 2 (\frac{\vec{a} + \vec{b} + \vec{c}}{2}) - (\vec{a} + \vec{b} + \vec{c}) = (\vec{a} + \vec{b} + \vec{c}) - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$

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