Show that if the diagonal of a quadrilateral are equal and bisect each other at right angle then it is a square
Answers
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 JACOB = XBOX = COD = LOAD =
90°
Now, to prove ABCD is a square, we
have to prove that ABCD is
parallelogram, AB = BC = CD = AD, and
one of its interior angles is 90°.
Proof:
In AAOB and ACID,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
angle AOB = angle COD(Vertically opposite
angles)
∆OBA~=∆COD (SAS congruence
rule)
AB = CD (By CPCT) (1)
And, ZOAB = ZOCD (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 GOD and ACID,
AO = CO (Diagonals bisect each other)
ZAID = ZAID (Given that each is 90°)
OD = OD (Common)
AAD = ACOD (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 AADC and ABCD,
AD = BC (Already proved)
AC = BD (Given)
DC = CD (Common)
AADC = ABCD (SSS Congruence
rule)
ZADCO = ABCD (By CPCT)
However, ZADCO + ABCD = 180°
(Co-interior angles)
ZAC + ZAC = 180°
However, ZADCO + ABCD = 180°
(Co-interior angles)
LADC + ZADC = 180°
2 ADC = 180°
LADC = 90°
One of the interior angles of
quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is
parallelogram, AB = BC = CD = AD and
one of its interior angles is 90°.
Therefore, ABCD is a square.
Step-by-step explanation:
Explanation
______________________________
Given that
Let ABCD be a quadrilateral
It's iagonals 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]
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 a square.
__________________________
HenceProved!