帮我讲下这道题---```plain 2. Question Stem: 如图，在△ABC中，∠ACB = 30°，AC = 4，D为BC上的一个动点，以BD为直径的圆O与AB相切于点B，交AD于点E，则CE的最小值为_______． Mathematical Information: - Triangle ABC (△ABC) - Angle ACB (∠ACB) = 30° - Side length AC = 4 - D is a moving point on BC. - Circle O has BD as diameter. - Circle O is tangent to AB at point B. - Circle O intersects AD at point E. Question Asked: Find the minimum value of CE. Diagram Description: - Type: Geometric figure (Triangle ABC with a circle). - Points: A, B, C, D, E, O. - Lines: AB, BC, AC, AD, CE. Line segment BC is horizontal. Point D is on BC. Point O is on BC (and is the midpoint of BD). - Shapes: Triangle ABC, Circle with diameter BD and center O. - Relationships: - Triangle ABC is shown with vertices A, B, C. - Point D is located on the line segment BC. - A circle with diameter BD and center O is drawn. O is the midpoint of BD. - The line AB is tangent to the circle at point B. - The line AD intersects the circle at point E (and A appears to be outside the circle). - Point E is on the line segment AD and on the circle. - The angle at C, ∠ACB, is indicated in the problem text as 30°. - The length of AC is given as 4. - From the tangency condition (circle with diameter BD tangent to AB at B), the radius OB is perpendicular to the tangent line AB at the point of tangency B. Since O lies on BD which lies on BC, this means AB is perpendicular to BC. Thus, ∠ABC = 90°. ```