请做一下图片中题目---**1. Question Stem:** 如图, 已知四边形 ABCD 为等腰梯形, AD // BC, AB=CD, AD=√2, E 为 CD 中点, 连接 AE, 且 AE=2√3, ∠DAE=30°, 作 AE ⊥ AF 交 BC 于 F, 则 BF = ( ) **2. Chart/Diagram Description:** * **Type:** Geometric figure. * **Main Elements:** * **Shape:** A quadrilateral ABCD is depicted, which is described as an isosceles trapezoid. * **Points:** Vertices A, B, C, D of the trapezoid are shown. Point E is located on the side CD. Point F is located on the base BC. * **Lines/Segments:** * The four sides of the trapezoid: AB, BC, CD, DA. * An internal line segment AE connects vertex A to point E on CD. * An internal line segment AF connects vertex A to point F on BC. * **Relative Position and Direction:** * AD appears as the top base, parallel to BC, the bottom base. * AB and CD are the non-parallel sides. * E is located on CD, approximately in the middle. * F is located on BC, to the left of C. * The lines AE and AF originate from vertex A. * **Labels and Annotations:** All vertices and key points are labeled: A, B, C, D, E, F. * **No coordinate axes, scales, or legend are present.**