show that if the diagonal of a quadrilateral bisect each other at right angle,then it is a rhombus
Answers
Step-by-step explanation:
We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
∴ In ΔAOB and ΔAOD, we have
AO = AO
[Common]
OB = OD
[Given that O in the mid-point of BD]
∠AOB = ∠AOD
[Each = 90°]
ΔAOB ≌ ΔAOD
[SAS criteria]
Their corresponding parts are equal.
AB = AD...(1)Similarly,AB = BC...(2) BC = CD...(3) CD = AD...(4)
∴ From (1), (2), (3) and (4), we have AB = BC CD = DA
Thus, the quadrilateral ABCD is a rhombus.
Answer:-
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
★ Given that:-
OA = OC
OB = OD
∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
★ To show that:-
If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
★ To prove that:-
ABCD is parallelogram and AB = BC = CD = AD
★ Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB [SAS congruency]
Thus, AB = BC [CPCT]
★ Similarly we can prove,
BC = CD
CD = AD
AD = AB
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Hence Proved.
