Question

Show that the diagonals of a square are equal and bisect each other at right angles.

Answer 1

Step- 1: Prove that the diagonals of a square are equal in length:

Consider the square $ABCD$

The diagonal of the square are $AC$and $BD$ which intersect each other at $O$

In $△ABC$ and $△DCB$

$\angle ABC=\angle DCB$[$\because$All interior angles are of $90°$]

$DC=AB$ [Sides of square]

$BC=CB$ [Common side]

$\therefore △ABC\cong △DCB$ [By SAS congruence rule]

$\therefore AC=DB$ [ By CPCT] …….(i)

So the diagonals of a square are equal in length.

Step- 2: Prove that the diagonals of a square bisect each other:

In $△AOB$ and $△COD$

$\angle AOB=\angle COD$[Vertically opposite angle]

$\angle ABO=\angle CDO$ [Alternate interior angle]

$AB=CD$ [Since sides of square are equal]

$\therefore △AOB\cong △COD$ [By AAS congruence rule]

$\therefore AO=CO$ and $OB=OD$ (CPCT ) …..(ii)

Hence the diagonal of a square bisect each other.

Step- 3: Prove that the diagonals of a square bisect each other at right angles:

In $△AOB$ and $△COB$ as we had proved that diagonals bisect each other,

$\therefore AO=CO$

$AB=CB$ ( Since, sides of a square are equal)

$BO$ is the common side of both the triangle

$\therefore △AOB\cong △COB$ [By SSS congruence rule]

$\therefore \angle AOB=\angle COB$ [By CPCT] …..(iii)

However $\angle AOB+\angle COB=180°$ (Linear pair)

$\therefore 2\angle AOB=180°$

$\Rightarrow \angle AOB=90°$ ….(iv)

Using (i), (ii) and (iv) we see that the diagonals of a square are equal and bisect each other at right angles

Hence,it is proved that the diagonals of a square are equal and bisect each other at right angles.