《2025年滚动迁移中考总复习数学山西专版》

1. 如图,在$\triangle ABC$中,点E为AB边上的点,$DE// BC$交AC于点D,$EF// AC$交BC于点F,$\frac{AE}{BE}=\frac{2}{5},BF = 8$,则DE的长为 ( )

A. $\frac{16}{5}$
B. $\frac{16}{7}$
C. 2
D. 3

答案： A

2. 如图,在Rt$\triangle ABC$中,$\angle ACB = 90^{\circ},CD$是$\triangle ABC$的中线,作$AF\bot CD$于点F,作$FE// BC$交BD于点E.已知$AC = 10,CF = 6$.
(1)求AB的长.

(2)求EF的长.

答案： 解：
(1)$\because \angle ACB = 90^{\circ},CD$是$\triangle ABC$的中线，
$\therefore CD = AD=\frac{1}{2}AB.\therefore \angle DCA=\angle DAC$.
$\because AF\perp CD$于点$F$，$\therefore \angle AFC = 90^{\circ}$.
$\therefore \angle AFC=\angle BCA.\therefore \triangle CAF\sim\triangle ABC.\therefore \frac{CF}{AC}=\frac{AC}{AB}$.
$\because AC = 10,CF = 6,\therefore AB=\frac{50}{3}$.
(2)$\because \angle ACB = 90^{\circ},CD = AD=\frac{1}{2}AB,\therefore CD=\frac{25}{3}$.
$\therefore DF = CD - CF=\frac{25}{3}-6=\frac{7}{3}$.
在$Rt\triangle ABC$中，$BC=\sqrt{AB^{2}-AC^{2}}=\frac{40}{3}$.
$\because EF// BC,\therefore \angle FED=\angle B$.
$\because \angle FDE=\angle CDB,\therefore \triangle FDE\sim\triangle CDB$.
$\therefore \frac{EF}{BC}=\frac{DF}{DC}.\therefore EF=\frac{56}{15}$.