感谢大家提供各种别出心裁的题解!
答案提供者:
Tan Jia Wei Daren |
Tristan Chaang |
林汶良这次不证明圆内接四边形 设 OC 与 BF 交于 G ∠BOG = ∠CFG (直角) ∠OGB = ∠FGC (对顶角) ∆BOG ∽ ∆CFG (两角对应相等) OG/FG = BG/CG OG/BG = FG/CG = OF/BC OF/BC = 1/√5 OF = 6/√5 |
L Yeeseng XD 小编:简单又粗暴。但在初中考试中,这属于作弊行为 😂😂😂 | |