9.「2025山东济南月考，★★★」【问题背景】
(1)如图1,在Rt△ABC和Rt△ADE中,AB=AC,AD=AE,由已知可以得到:
①△∽△;②△≌△.
【尝试应用】
(2)如图2,在△ABC和△ADE中,∠ACB = ∠AED = 90°,∠ABC=∠ADE=30°,求证:△ACE∽△ABD.
【问题解决】
(3)如图3,在△ABC和△ADE中,∠BAC = ∠DAE = 90°,∠ABC=∠ADE=30°,AC与DE相交于点F,点D在BC上,$\frac{AD}{BD}=\sqrt{3}$,求$\frac{DF}{CF}$的值.

(1)ABC;ADE;ABD;ACE.(ABC 与 ADE 可以互换,ABD 与 ACE 可以互换)
详解:$\because △ ABC$ 和 $△ ADE$ 是等腰直角三角形,
$\therefore △ ABC∽ △ ADE,∠ BAC=∠ DAE$,
$\therefore ∠ BAC-∠ DAC=∠ DAE-∠ DAC$,
$\therefore ∠ BAD=∠ CAE$,
又$\because AB=AC,AD=AE,\therefore △ ABD≌ △ ACE$.
(2)证明:$\because ∠ ACB=∠ AED=90°,∠ ABC=∠ ADE=30°$,
$\therefore △ ABC∽ △ ADE,\therefore \frac{AC}{AE}=\frac{AB}{AD},∠ CAB=∠ EAD$,
$\therefore ∠ CAB-∠ EAB=∠ EAD-∠ EAB,\frac{AC}{AB}=\frac{AE}{AD}$,
$\therefore ∠ CAE=∠ BAD,\therefore △ ACE∽ △ ABD$.
(3)如图,连接 $CE$,
易得 $△ ABD∽ △ ACE$,
$\therefore ∠ ACE=∠ ABD=∠ ADE=30°$,
又$\because ∠ AFD=∠ EFC,\therefore △ ADF∽ △ ECF,\therefore \frac{DF}{CF}=\frac{AD}{CE}$,
$\because \frac{AD}{BD}=\sqrt{3},\therefore AD=\sqrt{3}BD,\therefore \frac{DF}{CF}=\frac{AD}{CE}=\frac{\sqrt{3}BD}{CE}=\sqrt{3}· \frac{BD}{CE}$,
在 $\mathrm{Rt}△ ABC$ 中,$∠ ABC=30°$,
$\therefore BC=2AC,\therefore AB=\sqrt{BC^2-AC^2}=\sqrt{3}AC$,
$\because △ ABD∽ △ ACE,\therefore \frac{BD}{CE}=\frac{AB}{AC}=\sqrt{3},\therefore \frac{DF}{CF}=\sqrt{3}×\sqrt{3}=3$.