$ABCD$ is a quadrilateral in which $AB = BC$ and $AD = CD$. Show that $BD$ bisects both the angles $ABC$ and $ADC$.

It is given that $\triangle ABC \cong \triangle FDE$. If $AB = 5 \, cm$,$\angle B = 40^{\circ}$,and $\angle A = 80^{\circ}$,then which of the following is true?

$ABCD$ is a quadrilateral such that $AB = AD$ and $CB = CD$. Prove that $AC$ is the perpendicular bisector of $BD$.

Point $P$ lies in the interior of $\Delta ABC$. Prove that $PB + PC < AB + AC$.

In $\Delta ABC$,$AB = 8 \, cm$ and $BC = 5 \, cm$,then $AC > \ldots \ldots \ldots cm$.

$ABC$ 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?