ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. ). To prove that ∠BAD = ∠CAD, a student proceeded as follows:
In Δ ABD and Δ ACD,
AB = AC (Given)
∠B = ∠C (because AB = AC)
And ∠ADB = ∠ADC
Therefore, Δ ABD Δ Δ ACD (AAS)
So, ∠BAD = ∠CAD (CPCT)

What is the defect in the above arguments?