If diagonals of rhombus are equal then prove that it is a square
Answers
AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 900.
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 900.
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
So, ΔAOB ≅ ΔCOD (SAS congruence rule)
Hence 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.
So, 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)
So, ΔAOD ≅ ΔCOD (SAS congruence rule)
Hence 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)
So, ΔADC ≅ ΔBCD (SSS Congruence rule)
Hence, ∠ADC = ∠BCD (By CPCT)
However, ∠ADC + ∠BCD = 1800 (Co-interior angles)
=> ∠ADC + ∠ADC = 1800
=> 2∠ADC = 1800
=> ∠ADC = 900
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 900. Therefore, ABCD is a square.
AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 900.
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 900.
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
So, ΔAOB ≅ ΔCOD (SAS congruence rule)
Hence 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.
So, 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)
So, ΔAOD ≅ ΔCOD (SAS congruence rule)
Hence 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)
So, ΔADC ≅ ΔBCD (SSS Congruence rule)
Hence, ∠ADC = ∠BCD (By CPCT)
However, ∠ADC + ∠BCD = 1800 (Co-interior angles)
=> ∠ADC + ∠ADC = 1800
=> 2∠ADC = 1800
=> ∠ADC = 900
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 900. Therefore, ABCD is a