Question

Prove using vectors : If the diagonals of a parallelogram are equal in length, then it is a rectangle.

Answer 1

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Given : Let ABCD is a parallelogram and diagonals are equal.

To prove : if diagonals are equal ABCD is a rectangle

Proof :Let $\overrightarrow{DA}=\overrightarrow{a}$

$\overrightarrow{DC}=\overrightarrow{b}$

then $\overrightarrow{DB}=\overrightarrow{a}+\overrightarrow{b}$

and $\overrightarrow{AC}=\overrightarrow{b}-\overrightarrow{a}$

Since $|\overrightarrow{DB}|=|\overrightarrow{AC}|$ [Given]

$|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{b}-\overrightarrow{a}{|}^2$

$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{b}-\overrightarrow{a})$

$\Rightarrow 4\overrightarrow{a}.\overrightarrow{b}=0$ $[cos\theta =0]$

$\overrightarrow{a}.\overrightarrow{}b=0$ $[\theta ={90}^{\circ }]$

means $\overrightarrow{a}\perp \overrightarrow{b}$ and $\angle ADC={90}^{\circ }$

$\therefore$ ABCD is a rectangle.