1.如图,在矩形ABCD中,点E为AD的中点,BD和CE相交于点F.如果DF=2,那么线段BF的长度为.

1.4

2.如图，已知 $ AD · AB = AF · AC $.求证：$ △ DEB ∽ △ FEC $.

证明: $\because AD · AB = AF · AC, \therefore \frac{AD}{AF} = \frac{AC}{AB}. \because ∠ A = ∠ A$,
$\therefore △ ADC ∽ △ AFB, \therefore ∠ C = ∠ B$. 又 $\because ∠ DEB = ∠ FEC$,
$\therefore △ DEB ∽ △ FEC$.

3.如图,在△ABC中,BD平分∠ABC交AC于点D,点E在边BC上,满足∠BAE=∠ACB.连接AE交BD于点F,过点F作FG//BC交CD于点G.
(1)求证:AF=AD;
(2)求证:△DFG∽△FBA;
(3)若BE=2EF,求$\frac{DG}{FG}$的值.

(1)证明: $\because BD$ 平分 $∠ ABC, \therefore ∠ ABD = ∠ CBD. \because FG // BC$,
$\therefore ∠ FGA = ∠ ACB. \because ∠ BAE = ∠ ACB, \therefore ∠ BAE = ∠ ACB =$
$∠ FGA. \because ∠ AFD = ∠ BAE + ∠ ABD, ∠ ADB = ∠ CBD +$
$∠ ACB, \therefore ∠ AFD = ∠ ADB, \therefore AF = AD$.
(2)证明: $\because FG // BC, \therefore ∠ CBD = ∠ DFG, \therefore ∠ ABD =$
$∠ DFG, ∠ BAE = ∠ FGD, \therefore △ DFG ∽ △ FBA$.
(3)解: $\because ∠ BFE = ∠ AFD = ∠ ADF, ∠ ABD = ∠ EBD$,
$\therefore △ BEF ∽ △ BAD, \therefore \frac{EF}{BE} = \frac{AD}{BA} = \frac{1}{2}$, 又由(1)可得 $AD =$
$AF, \therefore \frac{AF}{BA} = \frac{1}{2}$. 由(2)知 $△ DFG ∽ △ FBA$, 故 $\frac{DG}{FG} = \frac{FA}{BA} = \frac{1}{2}$.

4.如图,在$△ ABC$中,$∠ ACB=90°$,$∠ A=30°$,将$△ ABC$绕点$C$顺时针旋转得到$△ A'B'C$,点$B'$在$AB$上,$A'B'$交$AC$于$F$,则图中与$△ AB'F$相似的三角形有(不再添加其他线段) ()


A.1个
B.2个
C.3个
D.4个

4.D

5.如图，在$△ ABC$中，边$AB$绕点$B$顺时针旋转$60°$与$BC$重合，点$D$，$E$分别在边$BC$，$AC$上，$∠ ADE=60°$。
(1)求证:$△ ABD ∽ △ DCE$；
(2)若$BD=2$，$CE=\dfrac{4}{3}$，求$△ ABC$的边长。

(1)证明: $\because$ 在 $△ ABC$ 中, 边 $AB$ 绕点 $B$ 顺时针旋转 $60°$ 与 $BC$ 重合, $\therefore △ ABC$ 为等边三角形, $∠ ABC = ∠ ACB = 60°, AB =$
$BC, \therefore ∠ BAD + ∠ BDA = 120°. \because ∠ ADE = 60°, \therefore ∠ BDA +$
$∠ CDE = 120°, \therefore ∠ BAD = ∠ CDE, \therefore △ ABD ∽ △ DCE$.
(2)解: $\because △ ABD ∽ △ DCE, \therefore AB: CD = BD: CE. \because BD = 2$,
$CE = \frac{4}{3}, \therefore AB: (BC - BD) = BD: CE, \therefore AB: (AB - 2) =$
$2: \frac{4}{3}, \therefore AB = 6, \therefore △ ABC$ 的边长为 6.