Question

In the figure given below, two equal chords are AB and CD of a circle with the center O intersect at right angles at P. If M and N are midpoint of the chords AB and CD respectively, prove that NOMP is a square.

Answer 1

Hint: The circle is a basic shape in $2D$ which is measured in the term of its radius.

A straight line divides the circle in the two-part and both points of that straight line lie on the circle.

The midpoint of the line divides it into two equal parts. The center of the circle has an equal distance from each point that lies on the circle.

All sides of the square have equal lengths. 

Complete step by step solution:

In the given figure  AB and CD are two chords and the midpoint of both chords are M and N respectively

The midpoint of chord AB $=$M

And the midpoint of the chord CD $=$N

Centre of the circle $=$O

Then we can say that 

$\Rightarrow \angle OMB=\angle OND={{90}^{0}}$

$\angle OMP=\angle ONP={{90}^{0}}$

Thus, the equal chords of the circle are equidistant from the circle, therefore

$\Rightarrow OM=ON$

Let’s see in the triangle OMP and ONP

$\Rightarrow OM=ON$

$\Rightarrow \angle OMP=\angle ONP$(Both ${{90}^{0}}$)

$\Rightarrow OP=OP$(Common side)

By side angle side rule $\Delta OMP\cong \Delta ONP$

$\Rightarrow MP=NP$

Thus in quadrilateral NOMP

$\Rightarrow OM=ON$

$\Rightarrow MP=NP$ and

$\Rightarrow \angle OMP=\angle ONP={{90}^{0}}$

Hence NOMP is square.

Note: 

1) There are some property of the square

2) It is a closed shape with 4 equal sides.

3) The square has 4 vertices and 4 sides.

4) The sum of all interior angles is equal to $360^{\circ}$.