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

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AB = AD$ and $CB = CD$.
To prove: $AC$ is the perpendicular bisector of $BD$.
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$AB = AD$ (Given)
$BC = CD$ (Given)
$AC = AC$ (Common side)
So,by $SSS$ congruence criterion,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$\angle 1 = \angle 2$ (by $CPCT$)
Now,in $\triangle AOB$ and $\triangle AOD$,we have:
$AB = AD$ (Given)
$\angle 1 = \angle 2$ (Proved above)
$AO = AO$ (Common side)
So,by $SAS$ congruence criterion,we have:
$\triangle AOB \cong \triangle AOD$
Therefore,$BO = DO$ (by $CPCT$)
And $\angle 3 = \angle 4$ (by $CPCT$)
But,$\angle 3 + \angle 4 = 180^{\circ}$ (Linear pair axiom)
Since $\angle 3 = \angle 4$,we have:
$\angle 3 + \angle 3 = 180^{\circ}$
$2\angle 3 = 180^{\circ}$
$\angle 3 = 90^{\circ}$
Since $BO = DO$ and $\angle 3 = 90^{\circ}$,$AC$ is the perpendicular bisector of $BD$.
Hence,proved.