这题怎么解，用上海初三年级水平所学的知识解答---23. (本题满分 12 分, 第(1)小题满分 5 分, 第(2)小题满分 7 分) 已知: 如图, 在梯形 $ABCD$ 中, $AD // BC$, $AB = CD$, $BD = BC$, $\angle DBC$ 的平分线交 $AD$ 延长线于点 $E$, 交 $CD$ 于点 $F$. (1) 求证: 四边形 $BCED$ 是菱形; (2) 连接 $AC$ 交 $BF$ 于点 $G$, 如果 $AC \perp CE$, 求证: $AB^2 = AG \cdot AC$. **Chart/Diagram Description:** * **Type:** Geometric figure (trapezoid and related lines). * **Main Elements:** * **Points:** A, B, C, D, E, F, G. * **Lines:** * Segments forming the trapezoid ABCD: AB, BC, CD, AD. * Diagonal BD. * Line segment AC (intersecting BF at G). * Line BF (starting from B, passing through G and F). * Line segment CE (connecting C and E). * Line segment DE (part of line AD extended). * Line ADE (line passing through A, D, E in that order). * **Shapes:** * Trapezoid ABCD with AD parallel to BC. * Triangle BCD. * Triangle ABD. * Triangle BDE. * Quadrilateral BCED. * **Labels and Annotations:** Points are labeled A, B, C, D, E, F, G. * **Relative Position and Direction:** * ABCD forms a trapezoid with AD parallel to BC. * BD is a diagonal. * E is on the extension of AD. * BF is the angle bisector of angle DBC, intersecting CD at F and AD extended at E. * AC intersects BF at G. * CE is drawn. **Other Relevant Text/Annotations (Possible scratch work):** * $\because AB=CD, BD=BC$ * $\angle BDC = \angle DCB$ * $AD // BC$ * $\angle EDB = \angle DBC = \angle BDE$ * $\frac{DF}{FC} = \frac{DE}{BE} = \frac{BF}{BF}$ (Likely incorrect or incomplete proportionality statement) * $\therefore BD = BC, AD$ bisects $\angle DBC$ (Error in interpretation, BF bisects $\angle DBC$) * $\therefore CD \perp BE$, $DF=FC$ (Likely conclusions or intermediate steps) * $\therefore DE = BC = AD$ (Likely conclusion or intermediate step) * $\therefore$ 四边形 BCED 是菱形 (Conclusion for part 1) * $\because$ BCED 是菱形 (Premise based on part 1)