7. 如图,在△ABC中,∠BAC=90°,AD⊥BC于点D,E为AC的中点,ED的延长线交AB的延长线于点F.求证:$\frac{AB}{AC}=\frac{DF}{AF}$

7. 证明:$\because ∠ BAC=90°,AD⊥ BC$,
$\therefore ∠ BAC=∠ BDA$.
又$\because ∠ CBA=∠ ABD,\therefore △ ABC∽△ DBA$,
$\therefore \frac{AB}{DB}=\frac{AC}{DA}$,即$\frac{AB}{AC}=\frac{DB}{DA},∠ BAD=∠ C$.
$\because AD⊥ BC,E$为$AC$的中点,$\therefore DE=EC$,
$\therefore ∠ BDF=∠ CDE=∠ C$,
$\therefore ∠ BDF=∠ BAD$.
又$\because ∠ F=∠ F,\therefore △ DBF∽△ ADF$,
$\therefore \frac{DB}{AD}=\frac{DF}{AF}$,即$\frac{AB}{AC}=\frac{DF}{AF}$.

8. 如图，$∠ C=∠ D=90°$，且$AB⊥ BE$，B为CD的中点. 若$CD=6$，$DE=2$，则AC的长为.

8. 4.5

9. 如图,在四边形ABCD中,∠BAD=∠ACB=∠ACD=45°,DE//BC交AC于点E,BC=3.5,AE=3,AD=8,则AB的值为.

9. $\frac{28}{3}$

10. 如图,在$△ ABC$与$△ ADE$中,点D在边BC上,$∠ 1 = ∠ 2 = ∠ 3$.若$AB = 3$,$AD = 2$,$AC = \frac{5}{2}$,则AE的长为.

10. $\frac{5}{3}$

11. 如图，$∠ DAB = ∠ EAC$，$∠ ADE = ∠ ABC$.
求证：
(1)$△ ADE ∽ △ ABC$；
(2)$\frac{AD}{AE}=\frac{BD}{CE}$.

11. 证明:(1)$\because ∠ DAB=∠ EAC$,
$\therefore ∠ DAE=∠ BAC$.
$\because ∠ ADE=∠ ABC,\therefore △ ADE∽△ ABC$.
(2)由(1)知$△ ADE∽△ ABC$,
$\therefore \frac{AD}{AB}=\frac{AE}{AC}$,即$\frac{AD}{AE}=\frac{AB}{AC}$.
$\because ∠ DAB=∠ EAC,\therefore △ ADB∽△ AEC$,
$\therefore \frac{AD}{AE}=\frac{BD}{CE}$.