解析---**Question ID:** P4 **Question Stem:** An acute triangle △ABC is inscribed in circle O, and ∠ACB > 2∠ABC. The incenter of △ABC is I. K is the reflection of I about BC. The extension of BA and the extension of KC intersect at point D. A line is drawn through B parallel to CI, intersecting the minor arc BC of circle O at E (E ≠ B). A line is drawn through A parallel to BC, intersecting BE at F. Prove: If BF = CE, then FK = AD. **Geometric Configuration Description:** The problem describes a geometric configuration, not an explicit diagram. The elements and their relationships are: * **Triangle:** An acute triangle △ABC. * **Circumcircle:** Circle O passes through vertices A, B, C. * **Angle Condition:** The angle at C (∠ACB) is greater than twice the angle at B (∠ABC). * **Incenter:** Point I is the incenter of △ABC. * **Point K:** K is the point symmetric to I with respect to the line BC. * **Point D:** Point D is the intersection of the extended line segment BA and the extended line segment KC. * **Point E:** A line drawn from B parallel to CI intersects the circle O at point E, specifically on the minor arc BC (E is not B). * **Point F:** A line drawn from A parallel to BC intersects the line segment BE at point F. * **Condition for Proof:** The condition to assume is that the length of segment BF is equal to the length of segment CE (BF = CE). * **To Prove:** The statement to prove is that the length of segment FK is equal to the length of segment AD (FK = AD). **Given Numerical Data (Example/Verification):** * DA = 9.07 厘米 * KF = 9.07 厘米 * CE = 4.45 厘米 * BF = 4.45 厘米