Question

$ABCD$ is a rectangle in which diagonal $BD$ bisects $\angle B$. Show that $ABCD$ is a square.

Answer 1

In a rectangle $ABCD$, diagonal $BD$ bisects $\angle B.$

Join $AC$.

In $△BAD$ and $△BCD,$

$\Rightarrow$ $\angle ABD=\angle CBD$ [ Since $BD$ bisect $\angle B$ ]

$\Rightarrow$ $\angle A=\angle C$ [ Each angle of a rectangle is ${90}^o.$ ]

$\Rightarrow$ $BD=BD$ [ Common side ]

$\therefore$ $△BAD\cong △BCD$ [ By AAS congruence rule ]

$\Rightarrow$ $AB=BC$ and $AD=CD$ [ CPCT ] ---- ( 1 )

We know opposite sides of rectangle are equal.

$\therefore$ $AB=CD$ and $BC=AD$ ---- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $AB=BC=CD=DA$

$\therefore$ $ABCD$ is a square.