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