Question

show that if the diagnols of quadrilateral bisect each other at right angke then it is a rhombus​

Answer 1

Answer:

$\huge{\underline{\underline{\tt{\red{Question}}}}}$

show that if the diagnols of quadrilateral bisect each other at right angke then it is a rhombus

$\huge{\underline{\underline{\tt{\blue{Answer:}}}}}$

$\huge{\underline{\underline{\tt{\blue{Given}}}}}$

A quadrilateral ABCD whose diagonal AC and BD intersect at O such that OA=OC and OB=OD and AC ⊥BD.

$\huge{\underline{\underline{\tt{\blue{To\ prove}}}}}$

ABCD is a rhombus

$\huge{\underline{\underline{\tt{\red{proof}}}}}$

since the diagonal of quadrilateral ABCD bisect each other therefore ABCD is a parallelogram.

Now, in ∆AOD and ∆COD, we have

OA=OC(GIVEN)

∠AOD=∠COD=90° [AC⊥BD]

and OD=OD (Common)

∆AOD≅∆COD(SAS RULE)

And So,AD=CD (C.P.CT)

Now, AB=CD and AD=BC (OPP. SIDES OF ||gm)

and AD=CD(PROVED)

BCOZ, AB=CD=AD=BC.

Hence,

$\huge{\underline{\underline{\tt{\blue{ABCD\ is\ a\ Rhombus}}}}}$