ABCD is a rectangle in which BD bisect $\angle$B & $\angle$D. Which of the following is correct option?

The correct option is A ABCD is square
In $△ABD$ & $△BDC$, we have
$\angle ADB=\angle BDC$ $[\because \angle D$ is bisected $]$
$\angle ABD=\angle DBC$ $[\because \angle B$ is bisected $]$
$BD=BD$ (common)
$\therefore △ABD\cong △BDC$ (By ASA rule)
$⟹AD=DC$ & $BC=AB$ (By CPCT) -------(1)

$AB=CD$ and $BC=AD$ (ABCD is a rectangle) -------(2)

From (1) and (2)
$\therefore AB=BC=CD=DA$ and $\angle A=\angle B=\angle C=\angle D={90}^{\circ }$ (ABCD is a rectangle)

$⟹$ ABCD is a square