ABCD is a rectangle in which diagonal AC bisects `angleA` as well as `angleC.` Show that
(i) ABCD is a square
(ii) diagonal AD bisects `angleB`

Explanation

ABCD is a reactangle.
`therefore" "AB=CDandBC=AD" "...(1)`
To prove : (1) ABCD is a square.
`i.e.," "AB=BC=CD=DA`
(ii) Diagonal AD bisects `angleB` as well as `angleD.`
Proof : (i) In `DeltaABCD and DeltaABC,` transversal interects
`because{:{(angleDAC=angleBAC,("given")),(angleDCA=angleBCA,("given")),(AC=CA,("common")):}`
`therefore" "DeltaADC~=DeltaABC" "("by ASA rule")`
`therefore" "AD=AB" "(c.p.c.t.)`
`and" "CD=BC" "...(2)`
Hence, from eqs. (1) and (2), we get
`" "AB=BC=AD=CD`
`therefore" "ABCD` is a square.
(ii) In `DeltaAOB and DeltaCOB`
`because{:{(AB=BC,("side of square")),(BO=OB,("common")),(OA=OC,(because"diagonals of square bisect each other")):}`
`therefore" "DeltaAOB~=DeltaCOB" "("by SSS rule")`
`therefore" "angleOBA=angleOBC" "(c.p.c.t)`
This shows that BO or BD bisect `angleB.`
Similarly, in `DeltaAOD and DeltaCOD`
`because{:{(AD=CD,("side of square")),(OD=DO,("common")),(OA=OC,(because"diagonals of square bisect each other")):}`
`therefore" "DeltaAOD~=DeltaCOD" "("by SSS rule")`
`therefore" "angleADO=angleCDO`
This shows that DO or DB bisect `angleD.`