Question

Let $\overrightarrow{a,}\overrightarrow{b,}\overrightarrow{c,}\overrightarrow{d}$be the position vectors of the four distinct points A, B, C, D. If $\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}$, then show that ABCD is a parallelogram.

Answer 1

Given: $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{d}$ are the position vectors of the four distinct points $A,B,C$ and $D$.
Also, we have, $\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}.$
$\Rightarrow \overrightarrow{AB}=\overrightarrow{DC}$
Again,
$$\begin{gathered} \overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d} \\ \Rightarrow \overrightarrow{b}-\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{d} \\ \Rightarrow \overrightarrow{CB}=\overrightarrow{DA} \end{gathered}$$
Consequently, $AB\parallel DC,CB\parallel DA$ and $\overrightarrow{AB}=\overrightarrow{DC},\overrightarrow{CB}=\overrightarrow{DA}$. Thus two of its opposite sides are equal and parallel.
Hence, $ABCD$ is a parallelogram.