52. Exemplar exercise problems XVII

\(ABC\) 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?

 

 

[Hint: Recall how \(\angle B = \angle C\) is proved when \(AB = AC\)].