In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.

If $\triangle ABC \cong \triangle PQR$ and $\triangle ABC$ is not congruent to $\triangle RPQ$,then which of the following is not true?

In $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.

In the figure,two lines $AB$ and $CD$ intersect each other at the point $O$ such that $BC \parallel DA$ and $BC = DA$. Show that $O$ is the midpoint of both the line segments $AB$ and $CD$.

If $AB = QR$,$BC = PR$,and $CA = PQ$,then

$ABC$ 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?