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\triangle ABF \cong \triangle AED\therefore \angle 2=\angle BAF\because \angle 3+\angle FAB=90^{\circ}\therefore \angle 2+\angle 3=90^{\circ}- 画出外接隐藏圆，以
O为圆心
- 套路，满满的套路！找出一个三角形吧：
\triangle OGB!,BG>=OB-OG!
