Question

show that if the diagonals of a quadrilateral are equal and bisect each other at right angle then it is a square

Answer 1

Let ABCD be the quadrilateral and AC and BD are it's Diagnol bisecting at O
Now in Triangle AOD and BOC.
AO= OC (given)
OB= OD (given)
angle AOD = angle BOC (each 90°.)
Hence triangles are congruent (by SAS rule)
By CPCT AD = BC
Now in Triangle ACD and BCD
CD= CD (common)
AC= BD (given)
AD= BC (proved above)
Hence triangle congruent by SSS rule
By CPCT angle C = D. =>[1]
Now angleC+ ang.D= 180 (Cointerior angle)
From [1],. 2( ang.C)=180°
Hence angle C and D = 90°
this proves that it is a rectangle
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Answer 2

Answer:

Given,

Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.

To prove,

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)

Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.

Thus, AB = CD by 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)

Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.

Thus, AD = CD by CPCT. --- (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB --- (ii)

also, ∠ADC = ∠BCD by CPCT.

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

∠ADC = 90° --- (iii)

One of the interior ang is right angle.Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.