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प्रश्न
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
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उत्तर
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In ΔAOD and ΔCOD,
OA = OC (Diagonals bisect each other)
∠AOD = ∠COD (Given)
OD = OD (Common)
∴ ΔAOD ≅ ΔCOD (By SAS congruence rule)
∴ AD = CD (1)
Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2),
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
संबंधित प्रश्न
In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.
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In a parallelogram, the diagonals bisect each other.
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