Consider a ΔABC with D, E and F being the midpoints of sides BC, CA and AB respectively. Let the position vectors of these vertices and midpoints be as shown in the figure.
We need to prove \(\vec {AD}+\vec {BE}+\vec{CF}=\vec{0}\)
As D is the midpoint of BC, using midpoint formula, we have
Similarly,
Recall the vector \(\vec {AD}\) is given by
\(\vec {AD}\) = position vector of D - position vector of A
But
\(\therefore\vec{AD}+\vec{BE}+\vec {CF}=\vec 0\)
Thus, the sum of the three vectors determined by the medians of a triangle is zero.
