Let D, E, F be the midpoints of the sides BC, | vedclass

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $D, E, F$ are the midpoints of sides $BC, CA, AB$

Vedclass · Accepted Answer

$\vec{0}$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $D, E, F$ are the midpoints of sides $BC, CA, AB$,their position vectors are:$\vec{D} = \frac{\vec{b} + \vec{c}}{2}$,$\vec{E} = \frac{\vec{c} + \vec{a}}{2}$,$\vec{F} = \frac{\vec{a} + \vec{b}}{2}$.Now,calculate the sum of the vectors:$\vec{AD} + \vec{BE} + \vec{CF} = (\vec{D} - \vec{A}) + (\vec{E} - \vec{B}) + (\vec{F} - \vec{C})$$= (\frac{\vec{b} + \vec{c}}{2} - \vec{a}) + (\frac{\vec{c} + \vec{a}}{2} - \vec{b}) + (\frac{\vec{a} + \vec{b}}{2} - \vec{c})$$= (\frac{\vec{b} + \vec{c} + \vec{c} + \vec{a} + \vec{a} + \vec{b}}{2}) - (\vec{a} + \vec{b} + \vec{c})$$= (\frac{2\vec{a} + 2\vec{b} + 2\vec{c}}{2}) - (\vec{a} + \vec{b} + \vec{c})$$= (\vec{a} + \vec{b} + \vec{c}) - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$.

Let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively. Then $\vec{AD} + \vec{BE} + \vec{CF} = \dots$

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