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Question
Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.
Solution
ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.
Now,
\[\vec{PQ} = \vec{PB} + \vec{BQ} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{BC} = \frac{1}{2}\left( \vec{AB} + \vec{BC} \right) = \frac{1}{2} \vec{AC}\]..............( 1 )
\[\vec{SR} = \vec{SD} + \vec{DR} = \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{DC} = \frac{1}{2}\left( \vec{AD} + \vec{DC} \right) = \frac{1}{2} \vec{AC}\] ...............( 2 )
From (1) and (2), we have
\[\vec{PQ} = \vec{SR}\]
So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.
Now,
\[\left| \vec{PQ} \right|^2 = \vec{PQ} . \vec{PQ} \]
\[ \Rightarrow \left| \vec{PQ} \right|^2 = \left( \vec{PB} + \vec{BQ} \right) . \left( \vec{PB} + \vec{BQ} \right)\]
\[ \Rightarrow \left| \vec{PQ} \right|^2 = \left| \vec{PB} \right|^2 + 2 \vec{PB} . \vec{BQ} + \left| \vec{BQ} \right|^2 \]
\[ \Rightarrow \left| \vec{PQ} \right|^2 = \left| \vec{PB} \right|^2 + 0 + \left| \vec{BQ} \right|^2 \left( \vec{PB} \perp \vec{BQ} \right)\]
\[ \Rightarrow \left| \vec{PQ} \right|^2 = \left| \vec{PB} \right|^2 + \left| \vec{BQ} \right|^2 ....................\left( 3 \right)\]
Also,
\[\left| \vec{PS} \right|^2 = \vec{PS} . \vec{PS} \]
\[ \Rightarrow \left| \vec{PS} \right|^2 = \left( \vec{PA} + \vec{AS} \right) . \left( \vec{PA} + \vec{AS} \right)\]
\[ \Rightarrow \left| \vec{PS} \right|^2 = \left| \vec{PA} \right|^2 + 2 \vec{PA} . \vec{AS} + \left| \vec{AS} \right|^2 \]
\[ \Rightarrow \left| \vec{PS} \right|^2 = \left| \vec{PB} \right|^2 + 0 + \left| \vec{BQ} \right|^2 \left( \vec{PA} \perp \vec{AS} \right)\]
\[ \Rightarrow \left| \vec{PS} \right|^2 = \left| \vec{PB} \right|^2 + \left| \vec{BQ} \right|^2........................\left( 4 \right)\]
From (3) and (4), we have
\[\left| \vec{PQ} \right|^2 = \left| \vec{PS} \right|^2 \]
\[ \Rightarrow \left| \vec{PQ} \right| = \left| \vec{PS} \right|\]
So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.
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