Question

plz solve this ????​

Answer 1

Answer:-

Given:- A quadrilateral ABCD whose diagonal AC and BD insert intersect at O such that OA is equal to AC and AB is equals to OD and AC perpendicular to BD.

To prove:- ABCD is a rhombus.

proof:- since the diagonal of a quadrilateral ABCD bisect each other, therefore ABCD is a parallelogram.

Now in triangle AOD and triangle COD we have

OA = OC (GIVEN)

AOD=COD=90 (AC perpendicular BD)

OD=OD (common)

TRIANGLE AOD congruent to TRIANGLE COD

ADis equals to CD (BY C.P.C.T)

AB= CD and AD=BC (opposite sides of a parallelogram)

AD=CD(proved)

AB=CD=AD=BC.

Hence ABCD is a Rhombus.

Hopefully it will help you