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

(N/A) We have a square $ABCD$ such that its diagonals $AC$ and $BD$ intersect at $O$.
$(i)$ To prove that the diagonals are equal,i.e.,$AC = BD$:
In $\Delta ABC$ and $\Delta BAD$,we have:
$AB = BA$ [Common]
$BC = AD$ [Opposite sides of the square $ABCD$]
$\angle ABC = \angle BAD = 90^{\circ}$ [All angles of a square are equal to $90^{\circ}$]
$\therefore \Delta ABC \cong \Delta BAD$ [$SAS$ criteria]
$\Rightarrow AC = BD$ ............. $(1)$
$(ii)$ To prove that $O$ is the mid-point of $AC$ and $BD$:
Because $AD \parallel BC$ and $AC$ is a transversal,$\angle 1 = \angle 3$ [Interior alternate angles].
Similarly,$\angle 2 = \angle 4$ [Interior alternate angles].
In $\Delta OAD$ and $\Delta OCB$,we have:
$AD = CB$ [Opposite sides of the square $ABCD$]
$\angle 1 = \angle 3$ [Proved]
$\angle 2 = \angle 4$ [Proved]
$\therefore \Delta OAD \cong \Delta OCB$ [$ASA$ criteria]
$\Rightarrow OA = OC$ and $OD = OB$
$\Rightarrow O$ is the mid-point of $AC$ and $BD$,i.e.,the diagonals $AC$ and $BD$ bisect each other at $O$. .......... $(2)$
$(iii)$ To prove that $AC \perp BD$:
In $\Delta OBA$ and $\Delta ODA$,we have:
$OB = OD$ [Proved]
$BA = DA$ [Opposite sides of the square]
$OA = OA$ [Common]
$\therefore \Delta OBA \cong \Delta ODA$ [$SSS$ criteria]
$\Rightarrow \angle AOB = \angle AOD$
Since $\angle AOB$ and $\angle AOD$ form a linear pair,$\angle AOB + \angle AOD = 180^{\circ}$.
$\Rightarrow \angle AOB = \angle AOD = 90^{\circ}$
$\Rightarrow AC \perp BD$ ............. $(3)$
From $(1)$,$(2)$,and $(3)$,we conclude that the diagonals of a square are equal and bisect each other at right angles.