Show that if daigonals of parralelogram …
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Sia ? 5 years, 2 months ago
Given: Diagonals of quadrilateral intersect each other at right angles.
To Prove: Quadrilateral is a rhombus.
Proof : In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}AOD,
AO = AO . . . [Common]
OB = OD . . . [Given]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}AOD . . .[Each 90o]
{tex}\therefore{/tex} {tex}\angle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}AOD . . . [By SAS property]
{tex}\therefore{/tex} AB = AD . . . [c.p.c.t.] . . . . (1)
Similarly, we can prove that
AB = BC . . . . (2)
BC = CD . . . . (3)
CD = AD . . . . (4)
From (1), (2), (3) and (4)
AB = BC = CD = DA
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.
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