Step-by-step explanation:Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and BISECT each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º. In ABC and DCB, AB = DC (Sides of a square are equal to each other) ABC = DCB (All interior angles are of 90) BC = CB (Common side) ABC = DCB (By SAS congruency) AC = DB (By CPCT) Hence, the diagonals of a square are equal in length. In AOB and COD, AOB = COD (Vertically opposite angles) ABO = CDO (Alternate interior angles) AB = CD (Sides of a square are always equal) AOB = COD (By AAS congruence rule) AO = CO and OB = OD (By CPCT) Hence, the diagonals of a square bisect each other. In  AOB and COB, As we had proved that diagonals bisect each other, therefore, AO = CO AB = CB (Sides of a square are equal) BO = BO (Common) AOB = COB (By SSS congruency) AOB = COB (By CPCT) However,AOB + COB = 180 (Linear pair) 2 AOB = 180º AOB = 90º Hence, the diagonals of a square bisect each other at right angles.