10. 如图，在$△ ABC$与$△ DCE$中，已知$∠ ACB=90°$，$∠ DCE=90°$，且$DC⊥ AB$，$DC$、$DE$分别交$AB$于点$M$、$N$。当$\frac{DN}{BC}=\frac{MN}{CM}$，$DE=10$时，求$CF$的长。

$CF=5$

11. 已知在$△ ABC$中，D、E是射线BC上的两点，且$BD=AB$，$CE=AC$.
(1)若$AB=AC$，且$∠ BAC=90°$(如图)，求证：$AE^2=BE· DE$；
(2)若$△ ABC$是直角三角形，且$AE^2=BE· DE$，求$∠ ABC$的度数.

11. (1) $\because AB = AC, ∠ BAC=90°$，
$\therefore ∠ B=∠ ACB=45°$.
$\because AC = CE, \therefore ∠ CAE = ∠ E=22.5°$,
$\because AB=BD,∠ B=45°$,
$\therefore ∠ BAD = ∠ ADB = \frac{180° -45°}{2}=67.5°$,
$\therefore ∠ DAC=90° -67.5°=22.5°$,
$\therefore ∠ DAE=22.5°+22.5°=45°$.
$\because ∠ DAE=∠ B,∠ E=∠ E$,
$\therefore △ EAD∽△ EBA,\frac{AE}{BE}=\frac{DE}{AE}$.
$\therefore AE^2=BE · DE$.
(2) $∠ ABC=45°$或$∠ ABC=30°$

如图,已知△ABC中,∠ACB=90°,AC=BC,点E、F在AB上,∠ECF=45°.
(1)求证:△ACF∽△BEC;
(2)设△ABC的面积为S,求证:AF·BE=2S.

(1) $\because ∠ ACB=90°,AC=BC$,
$\therefore ∠ A=∠ B=45°$.
$\because ∠ AFC = ∠ FCB + ∠ B =∠ FCB+45°$,
又 $∠ ECF = 45°, \therefore ∠ BCE =∠ ECF+∠ FCB=∠ FCB+45°$.
$\therefore ∠ AFC=∠ BCE$.
又 $\because ∠ A = ∠ B, \therefore △ ACF ∽ △ BEC$.
(2) $\because △ ACF∽△ BEC, \therefore \frac{AC}{BE}=\frac{AF}{BC}, \therefore BE · AF = AC · BC$. $\because S=\frac{1}{2}AC · BC, \therefore AC · BC = 2S$,
$\therefore AF · BE=2S$.