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 = 90o
Now, in ABC and DCB,
AB = DC (sides of square are equal)
ABC = DCB (all interior angles are of 90o)
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 = 180o (linear pair)
2AOB = 180 o AOB = 90o
Hence, the diagonals of a square bisect each other at right angles.