Question -
Answer -
Given that,
Let ABCD be a quadrilateral and its diagonals AC and BD bisecteach other at right angle at O.
To prove that,
The Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
, ΔAOB ≅ ΔCOD[SAS congruency]
Thus,
AB = CD [CPCT] — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB ||CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
, ΔAOD ≅ ΔCOD[SAS congruency]
Thus,
AD = CD [CPCT] — (ii)
also,
AD = BC and AD = CD
⇒ AD =BC = CD = AB — (ii)
also, ∠ADC = ∠BCD [CPCT]
and ∠ADC+∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒∠ADC =90° — (iii)
One of the interior angles is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is asquare.
Hence Proved.