6.「2025江苏无锡宜兴期中，★☆」如图，点E,F分别在正方形ABCD的边BC,CD上，且AE⊥EF.
(1)求证:△ABE∽△ECF.
(2)若BE=3,EC=7,求CF的长.

(1)证明:$\because$ 四边形 $ABCD$ 为正方形,
$\therefore ∠ B=∠ C=90°$,
$\because AE⊥ EF,\therefore ∠ AEF=90°$,
$\therefore ∠ BAE+∠ AEB=∠ AEB+∠ CEF=90°$,
$\therefore ∠ BAE=∠ CEF$,
$\therefore △ ABE∽ △ ECF$.
(2)$\because BE=3,EC=7$,
$\therefore AB=BC=BE+CE=10$,
$\because △ ABE∽ △ ECF$,
$\therefore AB:CE=BE:CF$,
$\therefore 10:7=3:CF,\therefore CF=2.1$.

7.「★★☆」如图,在四边形ABCD中,$AD=4$,$AB=10$,点E是AB的中点,连接DE,CE,若$∠ A=∠ B=∠ DEC$,求$\frac{BE}{BC}$的值.

$\because ∠ DEB$ 是 $△ ADE$ 的一个外角,
$\therefore ∠ DEB=∠ A+∠ ADE$,
$\because ∠ DEB=∠ DEC+∠ CEB,∠ A=∠ DEC$,
$\therefore ∠ ADE=∠ CEB$,
又$\because ∠ A=∠ B,\therefore △ DAE∽ △ EBC,\therefore \frac{AD}{BE}=\frac{AE}{BC}$,
$\because AB=10$,点 $E$ 是 $AB$ 的中点,
$\therefore AE=BE=5,\therefore \frac{BE}{BC}=\frac{AD}{AE}=\frac{4}{5}$.