Let $ABC$ be a triangle such that $AB=AC$. Suppose the tangent to the circumcircle of ABC at B is perpendicular to AC. Find angle ABC measured in degrees
1. Given that $ABC$ is an isosceles triangle with $AB = AC$.
2. Let the tangent to the circumcircle of $ABC$ at $B$ intersect $AC$ at $K$.
3. Since the tangent at $B$ is perpendicular to $AC$, we have $\angle KBA = 90^\circ$.
4. By the property of the tangent to a circle, $\angle KBA = \angle ACB$ (since the angle between the tangent and the chord through the point of tangency is equal to the angle in the alternate segment).
5. Let $\angle ACB = \theta$. Therefore, $\angle KBA = \theta$.
6. Since $AB = AC$, $\triangle ABC$ is isosceles, and $\angle ABC = \angle ACB = \theta$.
7. The sum of angles in $\triangle ABC$ is $180^\circ$. Therefore, $\angle BAC + 2\theta = 180^\circ$.
8. Since $\angle KAB = 2\theta$ (as $\angle KAB$ is the exterior angle for $\triangle KBA$), we have $\angle BAC = 2\theta$.
9. Substituting $\angle BAC = 2\theta$ into the angle sum equation, we get $2\theta + 2\theta = 180^\circ$.
10. Simplifying, $4\theta = 180^\circ$.
11. Therefore, $\theta = 45^\circ$.
12. Hence, $\angle ABC = \theta = 45^\circ$.

The final answer is $\boxed{45^\circ}$.