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


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 = 900



In ΔABC and ΔDCB,


AB = DC (Sides of a square are equal to each other)


ABC = DCB (All interior angles are of 90)


BC = CB (Common side)


ΔABC ΔDCB (By SAS congruency)


AC = DB (By CPCT)


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


In ΔAOB and ΔCOD,


AOB = COD (Vertically opposite angles)


ABO = CDO (Alternate interior angles)


AB = CD (Sides of a square are always equal)


ΔAOB ΔCOD (By AAS congruence rule)


AO = CO and


OB = OD (By CPCT)


Hence, the diagonals of a square bisect each other


In ΔAOB and ΔCOB,


As we had proved that diagonals bisect each other, therefore,


AO = CO


AB = CB (Sides of a square are equal)


BO = BO (Common)


ΔAOB ΔCOB (By SSS congruency)


AOB = COB (By CPCT)


However,


AOB + COB = 1800 (Linear pair)


2AOB = 1800


AOB = 900


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


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