show that the diagonals of a square are equal and bisect each other at right
Share
show that the diagonals of a square are equal and bisect each other at right
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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 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)
So, ΔABC ≅ ΔDCB (By SAS congruency)
Hence, 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)
So, ΔAOB ≅ ΔCOD (By AAS congruence rule)
Hence, 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)
So, ΔAOB ≅ ΔCOB (By SSS congruency)
Hence, ∠AOB = ∠COB (By CPCT)
However, ∠AOB + ∠COB = 180° (Linear pair)
2∠AOB = 180°
∠AOB = 90°
Hence, the diagonals of a square bisect each other at right angles.