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.

BTW. I LIKE THAT MEGAKHIGHT DP XD