Can we prove that $AD||PQ$ in the figure? In the given figure $AB=BC=CD$. If $PQ$ bisects $\angle APB$, then prove that $AD||PQ$.

My attempt: 

$$AB=BC=CD$$
  $$\angle APQ=\angle BPQ$$
  $$\text{arc } AB=\text{arc } BC$$
  $$\angle ADB=\angle BDC$$
  Thus we have $BD$ bisects $\angle ADC$. 

I am struck here. Please help me to complete the proof.
 A: Since $BC=CD$, we have
$$\angle BAC=\angle CAD= x$$
Using angles in the same segment,
$$\angle BAC=\angle BDC=x$$
You know that $\angle ADB=\angle BDC$, thus we have $\angle ADB=x$.
Since $\angle APB=\angle DPC=\angle CAD+\angle ADB$ (exterior angle of triangle), we now obtain $\angle APB=x+x=2x$.
Since $PQ$ bisects $\angle APB$, $\angle APQ=\frac{2x}{2}=x$.
Combining the results of $\angle APQ=\angle CAD= x$, we have proven that $AD$ is parallel to $PQ$ (alternate angles equal).  

The following picture is for reference.

A: By using your assumptions and the relations between arcs and angles in a circle, we do have the following facts:
$\mathrm{arc}AB = 2 \mathrm{arc} AQ' = \mathrm{arc} DC $, since we have $\angle APQ = \angle QPB$ and if we call the midpoint of $\mathrm{arc}AB$, $Q'$. Then we  have $\mathrm{arc} AQ' = \frac{\mathrm{arc} DC}{2}$. 
And since $\angle APQ = \mathrm{arc} AQ'  = \frac{\mathrm{arc} DC}{2} = \angle CAD. $ 
And it shows that two lines $AD$ and $PQ$ are parallel. 
And we are done.
A: Since $PA=PB$, so
$$\angle PAB=\angle PBA$$
Since $AB=BC$, so
$$\angle CAB = \angle ACB$$
Since $PB=PC$, so
$$\angle PCB = \angle PBC$$
Consider triangle ABC, we have
$$\angle CAB + \angle ABP + \angle CBP + \angle ACB = \pi$$
So
$$\angle PAB = \angle PBA = \pi/4$$
So,
$$\angle BPQ = \pi/4$$
Similarly, by consider triangle ADC, we have
$$\angle ADB = \pi/4$$
So by corresponding angles, the two lines are parallel.
