8. (2024·东海县期中)如图,A,C,E三点在同一条直线上,$∠ ACB=∠ E=90°$,$AB=AD$,$AE=$$CE+DE$.
(1)求证:$AC=DE$;
(2)求$∠ BAD$的度数.

8. (1)证明:$\because AE=AC+CE,AE=CE+DE$,
$\therefore AC+CE=CE+DE,\therefore AC=DE$.
(2)解:在$\mathrm{Rt}△ ABC$和$\mathrm{Rt}△ DAE$中,$\begin{cases} AB=DA,\\ AC=DE, \end{cases}$
$\therefore \mathrm{Rt}△ ABC≌\mathrm{Rt}△ DAE(\mathrm{HL})$,
$\therefore ∠ BAC=∠ D$.
$\because ∠ DAE+∠ D=90°,\therefore ∠ DAE+∠ BAC=90°$,
$\therefore ∠ BAD=90°$.

9. 如图,在$\mathrm{Rt}△ ABC$中,$∠ ACB=90°$,$CA=CB$,$D$是$AC$上一点,点$E$在$BC$的延长线上,且$AE=BD$,$BD$的延长线与$AE$交于点$F$. 试通过观察、测量、猜想等方法来探索$BF$与$AE$有何特殊的位置关系,并证明你的猜想.

9. 解:$BF⊥ AE$. 证明如下:
$\because ∠ ACB=90°,\therefore ∠ ACE=∠ BCD=90°$.
$\because BC=AC,BD=AE,\therefore \mathrm{Rt}△ BDC≌\mathrm{Rt}△ AEC(\mathrm{HL})$,
$\therefore ∠ CBD=∠ CAE$.
又$\because ∠ CAE+∠ E=90°,\therefore ∠ EBF+∠ E=90°$,
$\therefore ∠ BFE=90°$,即 $BF⊥ AE$.

10. 如图，$AC ⊥ BC$，$AD ⊥ BD$，$AD=BC$，$CE ⊥ AB$，$DF ⊥ AB$，垂足分别是 $E,F$.
求证:(1)$△ ABC ≌ △ BAD$;
(2)$CE=DF$.

10. 证明:(1)$\because AC⊥ BC,AD⊥ BD,\therefore ∠ ACB=∠ BDA=90°$.
在$\mathrm{Rt}△ ABC$和$\mathrm{Rt}△ BAD$中,$\begin{cases} AB=BA,\\ BC=AD, \end{cases}$
$\therefore \mathrm{Rt}△ ABC≌\mathrm{Rt}△ BAD(\mathrm{HL})$.
(2)$\because \mathrm{Rt}△ ABC≌\mathrm{Rt}△ BAD,\therefore S_{△ ABC}=S_{△ BAD}$.
$\because CE⊥ AB,DF⊥ AB,\therefore \frac{1}{2}AB· CE=\frac{1}{2}AB· DF$,
$\therefore CE=DF$.