In $\Delta ABC$,$AB = AC$. If $\angle A = 50^{\circ}$,then find $\angle B$ and $\angle C$.

$ABC$ 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?

Point $M$ lies on the perpendicular bisector of $PQ$. Also $M$ does not lie on $PQ$. Prove that $M$ is equidistant from $P$ and $Q$.

$D$ is a point on the side $BC$ of a $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then

Write the measures of its sides in ascending order in each of the following triangles:
$(1)$ In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 20^{\circ}$.

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?