ABC is an isosceles triangle with AB $=$AC and D is a point on BC such that $\mathit{AD}=\mathit{BC}$.

To prove that $\angle \mathit{BAD}=\angle \mathit{CAD},$ a student proceeded as follows-

In $\mathit{\Delta ABD}$and $\mathit{\Delta ACD},$

AB $=$AC (Given)

$\angle B=\angle C$(because AB $=$AC)

$\angle \mathit{ADB}=\angle \mathit{ADC}$

Then, $\mathit{\Delta ABD}\cong \mathit{\Delta ACD}$, (by ASS)

So, $\angle \mathit{BAD}=\angle \mathit{CAD}\left(\mathit{CPCT}\right)$

What is the defect in the above arguments?