Question

prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles,then it is a square.

Answer 1

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. 

Now, to prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.

Proof:

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite angles)

∴ ΔAOB ≅ ΔCOD (SAS congruence rule)

∴ AB = CD (By CPCT) ... (1)

And, ∠OAB = ∠OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.

∴ AB || CD ... (2)

From equations (1) and (2), we obtain

ABCD is a parallelogram.

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Given that each is 90º)

OD = OD (Common)

∴ ΔAOD ≅ ΔCOD (SAS congruence rule)

∴ AD = DC ... (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)

∴ AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

∴ ΔADC ≅ ΔBCD (SSS Congruence rule)

∴ ∠ADC = ∠BCD (By CPCT)

However, ∠ADC + ∠BCD = 180° (Co-interior angles)

⇒ ∠ADC + ∠ADC = 180°

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90°

One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90º. Therefore, ABCD is a square.

Answer 2

let a quadrilateral be ABCD and its diagonals intersect at point O forming right angles.since the diagonals are equal all the four bisected segments are also equal.
∴ OA = OB =OC =OD ---------- 1
now in the following figure, in ΔAOB and ΔAOD

AO=AO            ----------  common side
∠AOB= ∠AOD ----------   each equal to 90°
OB =OD          -----------   from 1
 ∴ by S-A-S criterion of congurency
ΔAOB  is congruent to Δ AOD

∴ AB=AD ---------- c.p.c.t

similarly 
AB = BC
BC =CD 
CD = AD 

finally, AB = BC = CD =AD --------------- 2
In ΔAOB 
OA=OB
∴∠OAB=∠OBA-----------angles opposite to equal sides are equal
 
Now, 
∠AOB+∠OAB+∠OBA=180°----------------- ANGLE SUM PROPERTY OF Δ
90° + 2∠OAB = 180°
∴∠OAB = ∠OBA = 45 °
similarly, 
∠ OBC = ∠OCB = 45°
∠OCD = ∠ODC = 45°
∠ODA = ∠OAD = 45°

Now, 
∠ABC=∠OBA + ∠OBC
          = 45°+45°
          =90°
similarly
∠BCD =90°
∠CDA = 90°
∠DAB = 90°
 
Now ,
∠ABC=∠BCD=∠CDA=∠DAB=90°-------------------3
 
From 2 and 3 all the angles are right angles and all the sides are equal 
∴ Quadrilateral ABCD is a SQUARE.