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

(N/A) Let $ABCD$ be a square. We need to prove that $AC = BD$,$OA = OC$,$OB = OD$,and $\angle AOB = 90^{\circ}$.
$1$. To prove diagonals are equal: Consider $\triangle ABC$ and $\triangle BAD$. In these triangles,$AB = BA$ (common side),$BC = AD$ (sides of a square),and $\angle ABC = \angle BAD = 90^{\circ}$. By $SAS$ congruence,$\triangle ABC \cong \triangle BAD$. Thus,$AC = BD$ $(CPCT)$.
$2$. To prove diagonals bisect each other: Consider $\triangle OAB$ and $\triangle OCD$. Here,$\angle OAB = \angle OCD$ (alternate interior angles),$\angle OBA = \angle ODC$ (alternate interior angles),and $AB = CD$ (sides of a square). By $ASA$ congruence,$\triangle OAB \cong \triangle OCD$. Thus,$OA = OC$ and $OB = OD$ $(CPCT)$.
$3$. To prove diagonals intersect at right angles: Consider $\triangle OAB$ and $\triangle OCB$. Here,$OA = OC$ (proved above),$AB = CB$ (sides of a square),and $OB = OB$ (common side). By $SSS$ congruence,$\triangle OAB \cong \triangle OCB$. Thus,$\angle AOB = \angle COB$ $(CPCT)$. Since $\angle AOB + \angle COB = 180^{\circ}$ (linear pair),$2 \angle AOB = 180^{\circ}$,which implies $\angle AOB = 90^{\circ}$.