In the given figure $AB=AC$ and $\angle ACD=105^{\circ }$, then $\angle BAC$ is equal to

1. A line segment extends from $B$ through $C$ to $D$. Thus, $\angle ACB$ and $\angle ACD$ form a linear pair:
$$\angle ACB+\angle ACD=180^{\circ }$$
$$\angle ACB+105^{\circ }=180^{\circ }⟹\angle ACB=75^{\circ }$$
In $△ABC$, we are given $AB=AC$. Therefore, it is an isosceles triangle where angles opposite to equal sides are equal:
$$\angle ABC=\angle ACB=75^{\circ }$$
Using the angle sum property of a triangle for $△ABC$:
$$\angle BAC+\angle ABC+\angle ACB=180^{\circ }$$
$$\angle BAC+75^{\circ }+75^{\circ }=180^{\circ }$$
$$\angle BAC+150^{\circ }=180^{\circ }⟹\angle BAC=30^{\circ }$$