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Question
Let \[\vec{a,} \vec{b,} \vec{c,} \vec{d}\] be the position vectors of the four distinct points A, B, C, D. If \[\vec{b} - \vec{a} = \vec{c} - \vec{d}\], then show that ABCD is a parallelogram.
Solution
Given: \[\vec{a} , \vec{b} , \vec{c}\] and \[\vec{d}\] are the position vectors of the four distinct points \[A, B, C\] and D Also, we have, \[\vec{b} - \vec{a} = c^\to - \vec{d} .\]
\[\Rightarrow \vec{AB} = \vec{DC}\]
Again,
\[\vec{b} - \vec{a} = \vec{c} - \vec{d} \]
\[ \Rightarrow \vec{b} - \vec{c} = \vec{a} - \vec{d} \]
\[ \Rightarrow \vec{CB} = \vec{DA}\]
Consequently,
\[AB \lVert DC , CB \lVert DA\]
\[\vec{AB} = \vec{DC} , \vec{CB} = \vec{DA}\]. Thus two of its opposite sides are equal and parallel.
Hence,
ABCD is a parallelogram.
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