In $\Delta PQR$,$PQ = PR$. If $\angle Q = 48^{\circ}$,then find $\angle P$ and $\angle R$.

$ABC$ 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?

$ABC$ is a right triangle such that $AB = AC$ and the bisector of angle $C$ intersects the side $AB$ at $D$. Prove that $AC + AD = BC$.

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

In the given figure,$AM$ and $BN$ are both perpendicular to $AB$. $MN$ intersects $AB$ at $P$. Also,$P$ is the midpoint of $AB$. Prove that $AM = BN$ and $P$ is the midpoint of $MN$.

If $\Delta ABC \cong \Delta RPQ,$ then which side of $\Delta PQR$ is equal to side $AB$ of $\Delta ABC$?