Question

If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, prove that $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{EF}$.

Answer 1

Let the position vectors of vertices A, B, C and D of quadrilateral ABCD be $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{d}$ respectively.

E is the mid point of AC

$\therefore$ Position vector of $E=\frac{1}{2}(\overrightarrow{a}+\overrightarrow{c})$

F is the mid point of BD

$\therefore$ Position vector of $F=\frac{1}{2}(\overrightarrow{b}+\overrightarrow{d})$

$\therefore \overrightarrow{EF}=$ position vector of F$=$position vector of $\hat{t}$

$=\frac{1}{2}(\overrightarrow{b}+\overrightarrow{d})-\frac{1}{2}(\overrightarrow{a}+\overrightarrow{c})$

$=\frac{1}{2}(\overrightarrow{b}+\overrightarrow{d}-\overrightarrow{a}-\overrightarrow{c})$

Now $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=(\overrightarrow{b}\cdot \overrightarrow{a})+(\overrightarrow{d}-\overrightarrow{c})+(\overrightarrow{d}-\overrightarrow{c})$

$=2(\overrightarrow{b}+\overrightarrow{d}-\overrightarrow{a}-\overrightarrow{c})$

$=2\times 2\overrightarrow{EF}$

$=4\overrightarrow{EF}$.