ABCD is a rectangle in which diagonal BD bisects `angle B`. then `ABCD` is a

Options

A. Rectangle
B. Rhombus
C. Parallelogram
D. Square

Answer

D

Explanation

Given In a rectangle ABCD, diagonal BD bisects `angleB`.
Construct Join AC.
To show ABCD is a square.

Proof
In `Delta`BAD and `Delta`BCD, `" "angleABD=angleCBD" "[given]`
`" "angleA=angleC" "["each "90^(@)]`
and `" "` BD=BD `" "` [common side]
`therefore" "DeltaBAD~=DeltaBCD" "`[by AAS congruence rule]
`therefore" "AB=BC`
and `" "AD=CD" "`[by CPCT rule]...(i)
But in rectangle ABCD, opposite sides are equal.
`therefore" "AB=CD`
and `" "BC=AD" "...(ii)`
From Eqs. (i) and (ii),
`" "AB=BC=CD=DA`
So, ABCD is a square. `" "` Hence proved.