In $\Delta ABC$,$AB = AC$ and $\angle CAD$ is an exterior angle of $\Delta ABC$ at vertex $A$. If ray $AP$ is the bisector of $\angle CAD$,then prove that $AP \parallel BC$.

(N/A) Given: In $\Delta ABC$,$AB = AC$. $\angle CAD$ is an exterior angle at vertex $A$. $AP$ is the bisector of $\angle CAD$.
To prove: $AP \parallel BC$.
Proof:
$1$. Since $AB = AC$,the angles opposite to equal sides are equal. Therefore,$\angle ABC = \angle ACB$.
$2$. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle CAD = \angle ABC + \angle ACB$.
$3$. Since $\angle ABC = \angle ACB$,we can write $\angle CAD = 2 \angle ACB$.
$4$. $AP$ is the bisector of $\angle CAD$,so $\angle CAP = \frac{1}{2} \angle CAD$.
$5$. Substituting the value from step $3$,$\angle CAP = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
$6$. $\angle CAP$ and $\angle ACB$ are alternate interior angles. Since they are equal,the lines $AP$ and $BC$ must be parallel $(AP \parallel BC)$.