$ABC$ is an isosceles triangle in which $AB = AC$. $AD$ bisects exterior angle $PAC$ and $CD \parallel AB$ (see Fig). Show that $ABCD$ is a parallelogram.

(N/A) Given: $AB = AC$ in $\triangle ABC$. Therefore,$\angle ABC = \angle ACB$.
Since $AD$ is the bisector of exterior angle $PAC$,we have $\angle PAD = \angle CAD = \frac{1}{2} \angle PAC$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle PAC = \angle ABC + \angle ACB$.
Since $\angle ABC = \angle ACB$,we have $\angle PAC = 2 \angle ACB$.
Substituting this into the bisector equation: $\angle CAD = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
These are alternate interior angles for lines $BC$ and $AD$ with transversal $AC$. Since the alternate interior angles are equal,$BC \parallel AD$.
We are given that $CD \parallel AB$.
Since both pairs of opposite sides are parallel ($BC \parallel AD$ and $AB \parallel CD$),$ABCD$ is a parallelogram.