if the diagonals of a quadrilateral bisect each other at right angles,then it is a rhombus
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Answer:
Step-by-step explanation:
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Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.Sol: 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.
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The diagonals bisect each other at right angles .
Check the attachment for the figure .
Let ABCD be the quadrilateral .
So O is the point of intersection of both diagonals .
In Δ AOB and Δ DOC ,
∠AOB = ∠DOC [ corresponding ∠s ]
BO = OD [ bisected diagonals ]
OA = OC [ bisected diagonals ]
Δ AOB ≅ Δ DOC [ S.A.S ]
AB = CD [ c.p.c.t ]
In Δ BOC and Δ AOB ,
OB = OB [ Common ]
AO = OC [ bisected diagonal ]
∠AOB = ∠AOC [ 90 each ]
Δ AOB ≅ Δ BOC [ S.A.S ]
AB = BC [ c.p.c.t ]
Hence :
AB = BC = CD
All sides must be equal .
Hence this has to be a rhombus .
