(A) True.
Let $EAF$ be the tangent to the circumcircle of $\triangle ABC$ at point $A$.
We need to prove that $EAF \parallel BC$.
By the alternate segment theorem,the angle between the tangent $EAF$ and the chord $AB$ is equal to the angle subtended by the chord $AB$ in the alternate segment,which is $\angle ACB$.
Thus,$\angle EAB = \angle ACB$.
Since $\triangle ABC$ is an isosceles triangle with $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$.
From these two relations,we get $\angle EAB = \angle ABC$.
Since these are alternate interior angles formed by the line $EAF$ and $BC$ with the transversal $AB$,the equality of these angles implies that $EAF \parallel BC$.