第10题怎么写？用上题目中的全部条件。用初中生的语言,条理清晰。---**Extraction Content:** **Question Stem:** 10. (2025·安徽蚌埠·三模) 如图, 在△ABC 中, ∠ACB = 90°, AC = BC, BD 是∠ABC 的平分线, 延长 DC 至点 E, 使得 CE = CD, 连接 BE, 过点 A 作 AF ⊥ BE 于点 F, 交 BD 于点 O, 交 BC 于点 H, 射线 CO 交 AB 于点 G, 连接 OE, CF, 则下列结论错误的是 ( ) **Options:** A. AC + CH = AB B. CG 是线段 AB 的垂直平分线 C. BH = OA D. BE = √2OE **Chart/Diagram Description:** * **Type:** Geometric figure (diagram of a triangle with several intersecting lines and points). * **Main Elements:** * A right-angled isosceles triangle ABC with ∠ACB = 90° and AC = BC. * Point D is on AC. * Line segment BD is the angle bisector of ∠ABC. * Point E is on the extension of line segment DC such that CE = CD. * Line segment BE is drawn. * Line segment AF is drawn from A, perpendicular to BE at point F. * Point O is the intersection of BD and AF. * Point H is the intersection of AF and BC. * Ray CO is drawn, intersecting AB at point G. * Line segments OE and CF are drawn. * Points are labeled as A, B, C, D, E, F, G, H, O. * Right angle symbol is shown at C (for ∠ACB) and at F (for ∠AFB or ∠AFE). * Lines and segments depicted include AC, BC, AB, BD, AD, CD, DE (extended), CE, BE, AF, BF, EF, OF, BH, CH, AH, CO, GO, CG, AG, BG, OE, CF.