Question

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

Answer 1

Given that $ABCD$ is a square.

To prove : $AC=BD$ and $AC$ and $BD$ bisect each other at right angles.

Proof:

(i) In a $\Delta ABC$ and $\Delta BAD$,

$AB=AB$ ( common line)

$BC=AD$ ( opppsite sides of a square)

$\angle ABC=\angle BAD$ ( = 90° )

$\Delta ABC\cong \Delta BAD$( By SAS property)

$AC=BD$ ( by CPCT).

(ii) In a $\Delta OAD$ and $\Delta OCB,$

$AD=CB$ ( opposite sides of a square)

$\angle OAD=\angle OCB$ ( transversal AC )

$\angle ODA=\angle OBC$ ( transversal BD )

$\Delta OAD\cong \Delta OCB$ (ASA property)

$OA=OC$ ---------(i)

Similarly $OB=OD$ ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a $\Delta OBA$ and $\Delta ODA$,

$OB=OD$ ( from (ii) )

$BA=DA$

$OA=OA$ ( common line )

$\Delta AOB=\Delta AOD$----(iii) ( by CPCT

$\angle AOB+\angle AOD=180°$ (linear pair)

$2\angle AOB=180°$

$\angle AOB=\angle AOD=90°$

∴$AC$ and $BD$ bisect each other at right angles.