In $\Delta ABC$ and $\Delta PQR$,$AB = PR$,$BC = RQ$ and $\angle B = \angle R$,then $\Delta ABC \cong \Delta \ldots$

$ABC$ 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?

In $\Delta ABC$,$AB = AC$ and $\angle B = 75^{\circ}$,then $\angle C = \dots$ (in $^{\circ}$)

$ABC$ and $DBC$ are two triangles on the same base $BC$ such that $A$ and $D$ lie on the opposite sides of $BC$,$AB = AC$ and $DB = DC$. Show that $AD$ is the perpendicular bisector of $BC$.

The line segment joining the mid-points $M$ and $N$ of the parallel sides $AB$ and $DC$ respectively of a trapezium $ABCD$ is perpendicular to both the sides $AB$ and $DC$. Prove that $AD = BC$.

The image of an object placed at a point $A$ before a plane mirror $LM$ is seen at the point $B$ by an observer at $D$ as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.