用初中二年级的知识讲解这个数学几何提 ---**Textual Information:** 如图，在Rt△ABC中，∠ACB = 90°，AE平分∠CAB交CB于点E，CD⊥AB于点D，交AE于点G，过点G作GF∥BC交AB于F，连接EF。 (1) 求证：CG = CE; (2) 判断四边形CGFE的形状，并证明; (3) 若AC = 3cm，BC = 4cm，求线段DG的长度。 **Chart/Diagram Description:** * **Type:** Geometric figure (a right-angled triangle with internal lines). * **Main Elements:** * A triangle ABC is shown. * The angle at vertex C is marked with a right-angle symbol, indicating ∠ACB = 90°. * A line segment AE is drawn from vertex A to side BC, intersecting BC at point E. AE appears to bisect ∠CAB. * A line segment CD is drawn from vertex C to side AB, intersecting AB at point D. A right-angle symbol is marked at D, indicating CD ⊥ AB. * Line segments AE and CD intersect at point G. * A line segment GF is drawn from point G to side AB, intersecting AB at point F. GF is shown as parallel to BC (implied by text, not explicitly marked with parallel symbols in this diagram). * A line segment EF is drawn connecting points E and F. * Points A, D, F, B lie on a straight line (hypotenuse AB). * Points B, E, C lie on a straight line (side BC). * Points A, G, E lie on a straight line (line segment AE). * Points C, G, D lie on a straight line (line segment CD).