Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Explanation

Given : A quadrialteral ABCD whose diagonals AC and BD bisect each other at right angles.
`i.e.," "OA=OCandOB=OD`
`and" "angleAOD=angleAOB=angleCOD=angleBOC=90^(@)`

To prove : ABCD is a rhombus.
Proof. In `DeltaOAB and DeltaODC,` we have
`because{:{(OA=OC,("given")),(OB=OD,("given")),(angleAOB=angleCOD,("vertically opposite angles")):}`
`therefore" "DeltaOAB~=DeltaOCD" "("by SAS rule")`
`therefore" "AB=CD" "("c.p.c.t")...(1)`
`because{:{(OA=OC,("given")),(OD=OB,("given")),(angleAOD=angleBOC,("vertically oposite angles")):}`
`therefore" "DeltaOAD~=DeltaOCB" "("by SAS rule")`
`therefore" "AD=BC" "(c.p.c.t.)...(2)`
Similarly, we can prove that
`" "{:(AB=AD),(CD=BC):}" "...(3)`
Hence, from (1), (2) and (3), we get
`" "AB=BC=AD=CD`
Hence, ABCD is a rhombus.