$ABC$ is an isosceles triangle with $AB = AC$ and $D$ is a point on $BC$ such that $AD \perp BC$. To prove that $\angle BAD = \angle CAD$,a student proceeded as follows:
In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (Given)
$\angle B = \angle C$ (because $AB = AC$)
and $\angle ADB = \angle ADC$
Therefore,$\triangle ABD \cong \triangle ACD$ $(AAS)$
So,$\angle BAD = \angle CAD$ $(CPCT)$
What is the defect in the above arguments?

(N/A) The defect in the student's argument is that they used the $AAS$ (Angle-Angle-Side) congruence criterion,but the side $AD$ is not included between the angles $\angle B$ and $\angle ADB$ in $\triangle ABD$.
Correct proof:
In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (Given)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^{\circ}$ (Given $AD \perp BC$)
Therefore,by $RHS$ (Right angle-Hypotenuse-Side) congruence criterion:
$\triangle ABD \cong \triangle ACD$
Thus,$\angle BAD = \angle CAD$ $(CPCT)$.