我是七年级的学生，给我讲讲这道题---**1. Question Stem:** 如图，在△ABC中，∠ABM=45°，AM⊥BM，垂足为M，点C是BM延长线上一点，连接AC.点D是线段AM上一点，MD=MC，点E是△ABC外一点，EC=AC，连接ED并延长交BC于点F，且点F是线段BC的中点，求证：∠BDF=∠CEF. **Translation/Interpretation of Question Stem:** As shown in the figure, in △ABC, ∠ABM=45°, AM⊥BM with the foot of the perpendicular at M. Point C is a point on the extension of BM. Connect AC. Point D is a point on line segment AM. MD=MC. Point E is a point outside △ABC. EC=AC. Connect ED and extend it to intersect BC at point F. Also, point F is the midpoint of line segment BC. Prove that ∠BDF=∠CEF. **Mathematical Formulas/Conditions:** * ∠ABM = 45° * AM ⊥ BM * MD = MC * EC = AC * F is the midpoint of BC **Task:** Prove: ∠BDF = ∠CEF **2. Chart/Diagram Description:** * **Type:** Geometric figure illustrating a triangle and related points and lines. * **Main Elements:** * **Points:** Labeled points A, B, C, D, E, F, M. * **Lines/Segments:** Segments connecting these points: AB, AC, BC, BM, MC, AM, DM, MD, ED, EF, EB, EC, DC, BD. * **Arrangement:** * Points B, F, M, C are collinear on a horizontal line, ordered from left to right. * Line segment AM is vertical, originating from M. * Point D is on the segment AM. * Point A is above D and M. * Point E is positioned to the right of the line BC and above C. * Segment ED is extended to pass through F, which lies on the line segment BC. * **Angles:** * ∠ABM is labeled or implied by the geometry. * ∠AMB is a right angle (90°) due to AM⊥BM. * **Labels and Annotations:** Points are labeled with letters A, B, C, D, E, F, M. The question number "1" is present at the top left. * **Relative Position and Direction:** AM is perpendicular to the line containing BM and MC at M. C is on the extension of BM, meaning M is between B and C. D is on AM, meaning D is between A and M or at A or M. F is on BC. E is outside △ABC. * **Legend:** None.