Question

plsss its a humble request to solve this fast​

Answer 1

In quadrilateral ABCD such that diagnol AC and BD bisect each other at right angle

In triangles AOB and AOD

OA= OA (common )

OB= OD (O is the mid point of BD )

angle AOB= angle AOD (90degree )

therefore ,triangles AOB=AOD (SAS)

their corresponding sides are

therefore ,AB= AD 1eq

AB= BC 2eq

BC= CD 3eq

CD= AD 4eq

therefore ,from 1,2,3,4 eq

AB= BC= CD= AD

therefore ,it is a rhombus