1. **Problem Statement:** Given a circle with center $O$ passing through points $A$, $B$, and $C$, and a tangent line $PQ$ touching the circle at $A$. It is given that $\angle PAB = \angle BAC$. We need to show that $PQ$ is tangent to the circle at $A$. 2. **Key Concept:** A line is tangent to a circle at a point if it is perpendicular to the radius drawn to that point. That is, if $PQ$ is tangent at $A$, then $OA \perp PQ$. 3. **Given:** $\angle PAB = \angle BAC$. Since $A$, $B$, and $C$ lie on the circle, $\angle BAC$ is an inscribed angle subtending arc $BC$. 4. **Step:** Consider triangle $PAB$. Since $\angle PAB = \angle BAC$, the line $AP$ is such that $\angle PAB$ equals the inscribed angle $\angle BAC$. 5. **Step:** The tangent at $A$ makes an angle with chord $AB$ equal to the angle in the alternate segment, which is $\angle BAC$. This is the Alternate Segment Theorem. 6. **Conclusion:** Since $\angle PAB = \angle BAC$, by the Alternate Segment Theorem, $PQ$ is tangent to the circle at $A$. **Final answer:** $PQ$ is tangent to the circle at point $A$ by the Alternate Segment Theorem.