这道题怎么做---```plain text (2) 【学以致用】 如图 2, 在等边△ABC中, 点D在BC边上 (CD<$\frac{1}{2}$BC), 点B关于直线AD的对称点为点E, 连接EC并延长交AD的延长线于点F, 连接BF. 1) 求∠AFE的度数 2) 求证: AF=BF+CF Diagram Description: * Type: Geometric figure. * Main Elements: * Points: A, B, C, D, E, F are labeled. * Shapes: Triangle ABC, Triangle BCD, Triangle ABD, Triangle BEF, Triangle CEF, Triangle ACF are formed by connecting the points. * Lines: Line segments AB, BC, CA form triangle ABC. Point D is on BC. Line segment AD is drawn. Point E is the reflection of B across AD. Line segment EC is extended to intersect the extension of AD at F. Line segment BF is drawn. * Relative Position: A is at the top, B is to the left, C is to the right, forming triangle ABC. D is on BC, closer to C than B. E is positioned such that it is the reflection of B across line AD. F is the intersection of the extension of EC and the extension of AD. * Labels: The figure is labeled "图 (2)". ```