Question

Prove by vector method, that a quadrilateral is a square if and only if diagonals are congruent and bisect each other at right angles.

Answer 1

Let ABCD be a quadrilateral.

This quadrilateral is square.

So,

$|\overrightarrow{AB}|=|\overrightarrow{BC}|=|\overrightarrow{CD}|=|\overrightarrow{DA}|$

Now,

$|\overrightarrow{CD}|=|\overrightarrow{DC}|$

$|\overrightarrow{DA}|=|\overrightarrow{AD}|$

The diagonal are $\overrightarrow{AC}and\overrightarrow{DB}$

It is square. Then the diagonals bisect each other.

We have to prove that : -

$\overrightarrow{AC}\perp \overrightarrow{DB}$

from square law of addition

We know that

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}and$

$\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}$

to prove that : - $\overrightarrow{AC}\perp \overrightarrow{DB}\Rightarrow \overrightarrow{AC}|.\overrightarrow{DB}=0$

$So,$

$\overrightarrow{AC}.\overrightarrow{DB}=(\overrightarrow{AB}+\overrightarrow{AD}).(\overrightarrow{AB}-\overrightarrow{AD})$

We know that,

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

$\Rightarrow \overrightarrow{AC}.\overrightarrow{DB}=|\overrightarrow{AC}{|}^2-|\overrightarrow{AD}{|}^2$

Since, ABCD is a square.

So,

$\Rightarrow |\overrightarrow{AB}|=|\overrightarrow{AD}|$

$\Rightarrow \overrightarrow{AC}|.\overrightarrow{DB}=0$

$\Rightarrow \overrightarrow{AC}\perp \overrightarrow{DB}$

Hence, Proved.