$ABC$ 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 $BC$.

(N/A) Given: $\triangle ABC$ and $\triangle DBC$ on the same base $BC$. Also,$AB = AC$ and $BD = CD$.
To prove: $AD$ is the perpendicular bisector of $BC$,i.e.,$OB = OC$ and $\angle AOB = \angle AOC = 90^{\circ}$.
Proof: In $\triangle ABD$ and $\triangle ACD$,we have:
$AB = AC$ [Given]
$BD = CD$ [Given]
$AD = AD$ [Common side]
So,by $SSS$ criterion of congruence,we have:
$\triangle ABD \cong \triangle ACD$
Therefore,$\angle 1 = \angle 2$ [$CPCT$]
Now,in $\triangle ABO$ and $\triangle ACO$,we have:
$AB = AC$ [Given]
$\angle 1 = \angle 2$ [Proved above]
$AO = AO$ [Common side]
So,by $SAS$ criterion of congruence,we have:
$\triangle ABO \cong \triangle ACO$
Therefore,$BO = CO$ [$CPCT$]
And,$\angle 3 = \angle 4$ [$CPCT$]
But,$\angle 3 + \angle 4 = 180^{\circ}$ [Linear pair axiom]
$\Rightarrow \angle 3 + \angle 3 = 180^{\circ}$
$\Rightarrow 2\angle 3 = 180^{\circ}$
$\Rightarrow \angle 3 = 90^{\circ}$
Since $BO = CO$ and $\angle 3 = 90^{\circ}$,$AD$ is the perpendicular bisector of $BC$. Hence,proved.