请给出解题步骤---**Extracted Content:** **Question Stem:** 1. 如图, 已知四边形 ABCD 为等腰梯形, AD∥BC, AB=CD, AD=$\sqrt{2}$, E 为 CD 中点, 连接 AE, 且 AE=$2\sqrt{3}$, ∠DAE=$30^\circ$, 作 AE⊥AF 交 BC 于 F, 则 BF= ( ) **Translation of Question Stem:** 1. As shown in the figure, it is known that quadrilateral ABCD is an isosceles trapezoid, AD∥BC, AB=CD, AD=$\sqrt{2}$, E is the midpoint of CD, connect AE, and AE=$2\sqrt{3}$, ∠DAE=$30^\circ$, make AE⊥AF intersecting BC at F, then BF= ( ) **Options:** A. 1 B. $3-\sqrt{3}$ C. $\sqrt{5}-1$ D. $4-2\sqrt{2}$ **Chart/Diagram Description:** * **Type:** Geometric figure, specifically an isosceles trapezoid ABCD with points E and F. * **Main Elements:** * **Shape:** Trapezoid ABCD. * **Vertices:** A, B, C, D are the vertices of the trapezoid. E is a point on CD. F is a point on BC. * **Lines:** Line segments AB, BC, CD, DA form the trapezoid. Line segments AE and AF are drawn from vertex A. Line segment DE and EC form CD. Line segment BF and FC form BC. * **Relative Position and Direction:** AD appears shorter than BC and is parallel to BC. AB and CD are the non-parallel sides. AE connects A to E on CD. AF connects A to F on BC. AE and AF appear to form a right angle at A. * **Labels:** Vertices are labeled A, B, C, D. Points E and F are labeled. * **Given Information (from text, not figure labels):** AD∥BC, AB=CD (isosceles trapezoid). AD = $\sqrt{2}$. E is the midpoint of CD. AE = $2\sqrt{3}$. ∠DAE = $30^\circ$. AE⊥AF (implies ∠EAF = $90^\circ$). F is the intersection of AF and BC. * **Not explicitly shown in figure but implied by text:** ∠EAF is a right angle.