CBSE Class 9 Answered

Let the diagonals AC and BD of the square ABCD intersect each other at a point O.

We need to prove AC = BD, OA = OC, OB = OD and AOB = 90^o

Now, in ABC and DCB,

AB = DC (sides of square are equal)

ABC = DCB (all interior angles are of 90^o)

BC = BC (common side)

ABC DCB (by SAS congruency)

AC = DB (by CPCT)

Hence, the diagonals of a square are equal in length.

Now, in AOB and COD,

AOB = COD (vertically opposite angles)

ABO = CDO (alternate interior angles)

AB = CD (sides of square are equal)

AOB COD (by AAS congruence rule)

AO = CO and OB = OD (by CPCT)

Hence, the diagonals of a square bisect each other.

Now in AOB and COB,

AO = CO

AB = CB (sides of square are equal)

BO = BO (common)

AOB COB (by SSS congruence)

AOB = COB (by CPCT)

But, AOB + COB = 180^o (linear pair)

2AOB = 180 ^o AOB = 90^o

Hence, the diagonals of a square bisect each other at right angles.