1. 如图，在$△ ABC$中，$D$，$E$分别是$AB$，$AC$上的点，且$BD=2AD$，$CE=2AE$. 求证：
(1)$△ ADE ∽ △ ABC$；
(2)$DF · BF=EF · CF$.

(1)$\because BD=2AD,CE=2AE$，
$\therefore AB=3AD,AC=3AE. \therefore \frac{AB}{AD}=\frac{AC}{AE}=3.$
又$\because ∠ A=∠ A,\therefore △ ADE ∽ △ ABC.$
(2)$\because △ ADE ∽ △ ABC,\therefore ∠ ADE=∠ ABC. \therefore DE// BC.$
$\therefore △ DEF ∽ △ CBF. \therefore \frac{DF}{CF}=\frac{EF}{BF},即 DF· BF=EF· CF.$

2. 如图,在$Rt△ ABC$中,$∠ BAC=90°$,$AB=AC$,M是边BC上的一个动点(不与点B,C重合),连接AM,$AN=AM$,且$∠ MAN=90°$,连接MN交AC于点P,连接CN.求证:
(1)$△ ABM≌△ ACN$;
(2)$MN^2=2AP· AB$.

(1)$\because ∠ BAC=90°,∠ MAN=90°,$
$\therefore ∠ BAM=∠ CAN.$
$\because AB=AC,AM=AN,\therefore △ ABM≌ △ ACN.$
(2)$\because AM=AN,∠ MAN=90°,\therefore ∠ ANM=45°,MN=\sqrt{2}AN.$
$\because ∠ BAC=90°,AB=AC,\therefore ∠ B=45°. \therefore ∠ B=∠ ANM.$
又$\because ∠ BAM=∠ CAN,\therefore △ ABM ∽ △ ANP.$
$\therefore \frac{AB}{AN}=\frac{AM}{AP}. \therefore AM· AN=AB· AP,即 AN^2=AB· AP.$
$\therefore MN^2=2AP· AB.$

3. 教材母题 教材P111复习题T5变式 如图,在$△ ABC$中,$∠ BAC=90°,AD⊥ BC$于点$D,E$是$AC$的中点,连接$ED$并延长,交$AB$的延长线于点$F$.求证:$\frac{AB}{AC}=\frac{DF}{AF}$.

$\because AD⊥ BC,\therefore ∠ ADB=∠ ADC=90°.$
在$Rt△ ADC$中,点E是斜边AC的中点，
$\therefore DE=CE=\frac{1}{2}AC. \therefore ∠ C=∠ CDE.$
又$\because ∠ CDE=∠ BDF,\therefore ∠ C=∠ BDF.$
$\because ∠ BAC=∠ BAD+∠ DAC=90°,∠ C+∠ DAC=90°,$
$\therefore ∠ BAD=∠ C=∠ BDF.$
又$\because ∠ F=∠ F,\therefore △ FDB ∽ △ FAD. \therefore \frac{DB}{AD}=\frac{FD}{FA}.$
$\because ∠ ADB=∠ ADC,∠ BAD=∠ C,$
$\therefore △ ABD ∽ △ CAD. \therefore \frac{BD}{AD}=\frac{AB}{AC}. \therefore \frac{AB}{AC}=\frac{DF}{AF}.$

4. 如图，AD是△ABC的高，DE⊥AB于E，DF⊥AC于F.求证:AE·AB=AF·AC.

$\because AD$是$△ ABC$的高,$DE⊥ AB,$
$\therefore ∠ AED=∠ ADB=∠ ADC=90°.$
$\because ∠ BAD=∠ EAD,\therefore △ AED ∽ △ ADB. \therefore \frac{EA}{AD}=\frac{AD}{AB}.$
$\therefore AD^2=AE· AB.$
同理可得$AD^2=AF· AC,\therefore AE· AB=AF· AC.$