Question

$AB$ and $CD$ are two equal chords of a circle with centre $O$ which intersect each other at right angle at point $P$. If $OM\perp AB$ and $ON\perp CD$ ; show that $OMPN$ is a square

Answer 1

$O$ is the centre.

Given:-$OM\perp AB$.....(1)

$ON\perp CD$......(2)

and $AB\perp CD$.....(3)

$\Rightarrow AB||ONandCD||OM$

from equations $\left(1\right),\left(2\right)$ and $\left(3\right)$

$\Rightarrow MP||ONandPN||OM$

(4) (5)

$\therefore OMPN$ is a parallelogram.

(since opposite side are parallel.)

But angle $\angle ONP={90}^0=\left(Given\right)$

$\angle NPM={90}^0\left(Given\right)$

$\angle NPM=\angle NOM$ [opposite angles of parallelogram]

$\therefore$ all angles in the parallelogram are ${90}^0$

$\therefore OMPN$ is a square or rectangle.

$\therefore$ length of perpendicular from centre to equal chords are equal.

$\Rightarrow ON=OM$

adjacent sides of rectangular $OMPN$ are equal.

$\Rightarrow OMPNisasquare$