已知 \(\Box ACBE\) 中，\(AB=AC\)，圆 \(O\) 的半径 \(r=\dfrac{5}{2}\)，\(BC=4\)。
作 \(OH\perp BC\)，根据垂径定理可得，\(BH=CH=2\)。
\(\therefore OH=\sqrt{{OB}^2-{BH}^2}=\dfrac{3}{2}\)，\(AH=AO+OH=\dfrac{5}{2}+\dfrac{3}{2}=4\)，\(AC=AB=\sqrt{{BH}^2+{AH}^2}=2\sqrt{5}\)
\(\because A,F,B,C\) 四点共圆，\(\therefore \angle AFB + \angle ACB = 180^{\circ}\)
\(\therefore \angle AFE = \angle ACB\)
又 \(\because \angle AEF = \angle ACB\)，\(\therefore \angle AEF = \angle AFE = \angle ABC = \angle ACB\)。
\(\therefore \triangle ABC \sim \triangle AEF\)。
\(\therefore \dfrac{AB}{AE}=\dfrac{BC}{EF}\)。
\(\therefore EF=\dfrac{AE}{AB}\times BC=\dfrac{4}{2\sqrt{5}}\times 4=\dfrac{8}{5}\sqrt{5},\ BF=BE-EF=\dfrac{2}{5}\sqrt{5}\)
标签：ACB,angle,BC,dfrac,sqrt,一道,therefore From： https://www.cnblogs.com/DaYanZaiHappy/p/17342175.html