过点

B

作

B J ⊥ E C

于点

J

，

$∵ A C$ 是直径，

$∴ ∠ A B C = 90 °$

$∴ \sin ∠ B A C = \frac{B C}{A C} = \frac{1}{3}$

设 $B C = a ， A C = 3 a$ ，

$∵ B J ⊥ A C$

$∴ ∠ A J B = 90 °$

$∴ ∠ B A C + ∠ A B J = 90 ° ， ∠ A B J + ∠ C B J = 90 °$

$∴ ∠ C B J = ∠ B A C$

$∴ \sin ∠ C B J = \sin ∠ B A C = \frac{C J}{B C} = \frac{1}{3}$

$∴ C J = \frac{1}{3} a$

$∴ B J = \sqrt{B C^{2} - C J^{2}} = \sqrt{a^{2} - {(\frac{1}{3} a)}^{2}} = \frac{2 \sqrt{2}}{3} a$

$∵ D A = D E$

$∴ ∠ D A E = ∠ A E D = ∠ C E B$

$∵ ∠ D A E = ∠ C B E$

$∴ ∠ C E B = ∠ C B E$

$∴ C E = C B = a$

$∴ E J = E C - C J = a - \frac{1}{3} a = \frac{2}{3} a ， A E = A C - E C = 2 a$

$∵ A F / / B J$

$∴ \frac{A F}{B J} = \frac{A E}{E J}$

$∴ \frac{2 \sqrt{6}}{\frac{2 \sqrt{2}}{3} a} = \frac{2 a}{\frac{2}{3} a}$

$∴ a = \sqrt{3}$

$∴ A E = 2 \sqrt{3} ， E J = \frac{2 \sqrt{3}}{3} ， B J = \frac{2 \sqrt{6}}{3}$

$∴ E F = \sqrt{A F^{2} + A E^{2}} = \sqrt{{(2 \sqrt{6})}^{2} + {(2 \sqrt{3})}^{2}} = 6 ， B E = \sqrt{E J^{2} + B J^{2}} = \sqrt{{(\frac{2 \sqrt{3}}{3})}^{2} + {(\frac{2 \sqrt{6}}{3})}^{2}} = 2$ $∴ B F = E F + B E = 8$ .

四川省成都市高新区2020-2021学年九年级下学期诊断性考试（3月）