针对这个道题目出讲解---完成下面的推理, 并在括号内标注理由: 如图, 点 D, E 分别在 AB, AC 上, 连接 DE, BE, 点 F 在 BE 上, 连接 DF, 已知 ∠EDF=∠C, ∠2+∠3=180°, 求证: ∠BDE+∠DBC=180°. 证明: ∵ ∠2+∠3=180° ( 已知), ∵ ∠3+∠4=180° ( ______) , ∴ ∠2=______ ( 同角的补角相等), ∴ DF//______ ( ______ ), ∴ ∠EDF=∠1 ( ______ ), ∵ ∠EDF=∠C (已知), ∴ ______ (等量代换), ∴ ______ (同位角相等, 两直线平行), ∴ ∠BDE+∠DBC=180° ( ______ ). **Chart/Diagram Description:** * Type: Geometric figure (triangle with internal lines and points). * Main Elements: * A triangle labeled ABC. * Point D is located on the side AB. * Point E is located on the side AC. * Line segment DE connects D and E. * Line segment BE connects B and E. * Point F is located on the line segment BE, between B and E. * Line segment DF connects D and F. * Angles are labeled with numbers: * Angle 1 (∠1) is formed by lines DE and BE, near point E. * Angle 2 (∠2) is formed by line BE and AC, near point E (likely ∠BEC). * Angle 3 (∠3) is formed by lines DF and BE, near point F (likely ∠DFB). * Angle 4 (∠4) is formed by lines DF and BE, near point F, adjacent to ∠3 (likely ∠DFE). ∠3 and ∠4 are adjacent angles on the straight line BE.