Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Let ABCD be a quadrilateral and its diagonals AC and BD bisect each 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]
AB = CD [CPCT] — (i)
∠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]
AD = CD [CPCT] — (ii)
AD = BC and AD = CD
⇒ AD = BC = CD = AB .......(ii)
∠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 a square.
Hence Proved.