10. 已知$△ ADE$的顶点$E$在$△ ABC$的内部，点$D$、点$E$在直线$AC$同侧.
(1)如图①，连接$BD$，$BE$，$CE$，若$△ ABC$和$△ ADE$是等边三角形，$C,D,E$三点共线，$CE:DE=1:2$，求$S_{△ ADE}:S_{△ ABC}$的值；
(2)如图②，连接$BD$，$BE$，$CE(C,D,E$三点不共线$)$，$∠ BAC=∠ DAE=n°(0＜n＜90)$，若$AB=AC$，$AD=AE$，求$∠ BEC-∠ DBE$的值(用含$n$的代数式表示)；
(3)若$△ ABC$是等腰三角形，$AB=BC=5$，$AC=8$，$BH⊥ AC$，点$E$在高$BH$上，点$D$在$HB$的延长线上，连接$AE$并延长交边$BC$于点$F$，连接$DF$，$DA$，当$∠ DAE=∠ ABH$，$△ ABD$与$△ BDF$相似时，$EH$的长为.

(1) $\because CE:DE=1:2$,$\therefore$ 设 $CE=k$,$DE=2k$.
$\because △ ADE$ 是等边三角形,
$\therefore AE=DE=2k$,
过点 $A$ 作 $AF⊥ CD$,则易得 $EF=\dfrac{1}{2}DE=k$,
$\therefore CF=CE+EF=2k$,$AF=\sqrt{AE^2-EF^2}=\sqrt{3}k$,
$\therefore AC=\sqrt{AF^2+CF^2}=\sqrt{7}k$.
$\because △ ADE$ 是等边三角形,$△ ABC$ 是等边三角形,
$\therefore \dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}$,$\therefore △ ADE ∽ △ ABC$,
$\therefore S_{△ ADE}:S_{△ ABC}=AE^2:AC^2=\dfrac{4}{7}$.
(2) $\because ∠ BAC = ∠ DAE = n°$,$\therefore ∠ BAD = ∠ CAE = n° - ∠ BAE$.
$\because AB=AC$,$AD=AE$,$\therefore △ ADB ≌ △ AEC$,
$\therefore ∠ ABD = ∠ ACE$.
在 $△ ABC$ 中,$∠ BAC + ∠ ABC + ∠ ACB = 180°$,
在 $△ BCE$ 中,$∠ BEC + ∠ EBC + ∠ BCE = 180°$,
$\therefore ∠ BAC + ∠ ABC + ∠ ACB = ∠ BEC + ∠ EBC + ∠ BCE$,
即 $n° + (∠ ABC - ∠ EBC) + (∠ ACB - ∠ BCE) = ∠ BEC$,
$\therefore n° + ∠ ABE + ∠ ACE = ∠ BEC$,
$\therefore n° + ∠ ABE + ∠ ABD = ∠ BEC$,
$\therefore n° + ∠ EBD = ∠ BEC$,$\therefore ∠ BEC - ∠ DBE = n°$.
(3)【点拨】$\because AB=BC=5$,$AC=8$,$BH⊥ AC$,
$\therefore AH=CH=4$,$∠ ABH = ∠ CBH$,
$\therefore BH$ 垂直平分 $AC$,$BH=\sqrt{AB^2-AH^2}=3$,
当 $∠ DAE = ∠ ABH$,$△ ABD$ 与 $△ BDF$ 相似时:
①当 $∠ ADB = ∠ FDB$ 时:
$\because ∠ ABH = ∠ CBH$,$\therefore ∠ ABD = ∠ CBD$.
$\because DB=DB$,$\therefore △ ADB ≌ △ FDB(\mathrm{ASA})$,$\therefore AD=FD$.
连接 $CD$,$\because$ 点 $D$ 在 $AC$ 的中垂线上,$\therefore AD=CD$.
$\therefore CD=FD$,
$\because$ 点 $F$ 在 $BC$ 上,$\therefore$ 点 $C,F$ 重合.
此时点 $E$ 与点 $H$ 重合(不合题意,舍去);
②如图,当 $∠ ADB = ∠ BFD(∠ DAB = ∠ BDF)$ 时:
$\because ∠ ABH = ∠ ADB + ∠ BAD$,$∠ DAE = ∠ BAE + ∠ BAD$,
$\therefore ∠ ADB = ∠ BAE$,
$\therefore ∠ DAE = ∠ ADF = ∠ ABH$,
$\therefore AF=DF$.
过点 $F$ 作 $FG⊥ AC$ 于点 $G$,$FK⊥ AD$ 于点 $K$,
$\therefore AD=2AK$.
$\because \cos ∠ DAF = \cos ∠ ABH$,$\therefore \dfrac{AK}{AF} = \dfrac{BH}{AB} = \dfrac{3}{5}$.
设 $AK=3x$,则 $AF=5x$,$AD=6x$,
$\therefore DF=AF=5x$.
$\because △ ABD$ 与 $△ BDF$ 相似,
$\therefore$ 易得 $\dfrac{AD}{DF} = \dfrac{AB}{DB} = \dfrac{DB}{FB}$,即 $\dfrac{6x}{5x} = \dfrac{5}{DB} = \dfrac{DB}{BF}$,
$\therefore DB=\dfrac{25}{6}$,$\therefore BF=\dfrac{125}{36}$,$\therefore CF=BC-BF=\dfrac{55}{36}$,
$\therefore FG=CF· \sin ∠ ACB = CF· \dfrac{BH}{CB} = \dfrac{55}{36}× \dfrac{3}{5} = \dfrac{11}{12}$,
$\therefore CG=\sqrt{CF^2-FG^2}=\dfrac{11}{9}$,$\therefore AG=AC-CG=\dfrac{61}{9}$.
$\because \tan ∠ FAG = \dfrac{EH}{AH} = \dfrac{FG}{AG} = \dfrac{\dfrac{11}{12}}{\dfrac{61}{9}}$,$AH=4$,$\therefore EH=\dfrac{33}{61}$.