Question

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

Answer 1

In a Rhombus $OACB$, Let $OA\uparrow =a\uparrow$ and $OB\uparrow =b\uparrow$

Then $OA=OB$ i.e, $a=b$......$\left(1\right)$

The diagonals of this Rhombus are $OC\uparrow$ and $BA\uparrow$

The P.V.S of points $O,A,C,B$ are $O\uparrow ,a\uparrow ,c\uparrow =(a\uparrow +b\uparrow )$ and $b\uparrow$

$\therefore$ mid pouint of $OC$ is $M(m\uparrow )=\frac{o\uparrow +c\uparrow }{2}=\frac{a\uparrow +b\uparrow }{2}$

$\therefore m\uparrow =n\uparrow \therefore M\equiv N$

$\therefore OC$ and $BA$ have the same mid point

$\therefore$ Diagonals bisect each other.

also

$OC\uparrow .BA\uparrow$

$=(a\uparrow +b\uparrow ).(a\uparrow -b\uparrow )$

$=(a\uparrow {)}^2-(b\uparrow {)}^2$

$=|a\uparrow {|}^2-|b\uparrow {|}^2$

$=a^2-b^2$ where $a=b$

$=0$

$\therefore$ Diagonals are mutually perpendicular.