答案：
1.(1)证明:$\because ∠BAC=∠DAE=90°$,
$\therefore ∠BAC-∠DAC=∠DAE-∠DAC$,
即$∠BAD=∠CAE$.
在$△ BAD$和$△ CAE$中,$\begin{cases} AB=AC, \\ ∠BAD=∠CAE, \\ AD=AE, \end{cases}$
$\therefore △ BAD≌△ CAE(\mathrm{SAS})$.
(2)解:$2AD^2=BD^2+CD^2$,理由:连接EC.如答图.
$\because ∠BAC=∠DAE=90°$,$\therefore ∠BAD=∠CAE$.
在$△ ABD$和$△ ACE$中,$\begin{cases} AB=AC, \\ ∠BAD=∠CAE, \\ AD=AE, \end{cases}$
$\therefore △ ABD≌△ ACE(\mathrm{SAS})$,
$\therefore BD=CE$,$∠B=∠ACE=45°$.
又$\because ∠ACB=∠B=45°$,
$\therefore ∠BCE=∠ACB+∠ACE=45°+45°=90°$,
$\therefore DE^2=CE^2+CD^2$.
在$\mathrm{Rt}△ ADE$中,$DE^2=AD^2+AE^2=2AD^2$,
$\therefore 2AD^2=BD^2+CD^2$.

第1题答图

答案：
2.(1)证明:$\because DE⊥ DF$,$\therefore ∠EDF=90°$.
$\because ∠BAC=90°$,$\therefore ∠AFD+∠AED=180°$.
$\because ∠BED+∠AED=180°$,$\therefore ∠BED=∠AFD$.
(2)证明:如答图①,延长ED到点P,使$DP=DE$,连接FP,CP.

第2题答图①
$\because D$是BC的中点,$\therefore BD=CD$.
在$△ BED$和$△ CPD$中,$\begin{cases} ED=PD, \\ ∠EDB=∠PDC, \\ BD=CD, \end{cases}$
$\therefore △ BED≌△ CPD(\mathrm{SAS})$,$\therefore BE=CP$,$∠B=∠PCD$.
在$△ EDF$和$△ PDF$中,$\begin{cases} DE=DP, \\ ∠EDF=∠PDF=90°, \\ DF=DF, \end{cases}$
$\therefore △ EDF≌△ PDF(\mathrm{SAS})$,$\therefore EF=FP$.
$\because ∠BAC=90°$,$\therefore ∠B+∠ACB=90°$,$\therefore ∠ACB+∠DCP=90°$,即$∠FCP=90°$.
在$\mathrm{Rt}△ FCP$中,根据勾股定理,得$CF^2+CP^2=PF^2$,
$\because BE=CP$,$PF=EF$,$\therefore BE^2+CF^2=EF^2$.
(3)解:如答图②,连接AD,过点D作$DG⊥ EF$于点G.
由题意知$△ ABC$为等腰直角三角形.
$\because D$为BC边的中点,$\therefore ∠BAD=∠FCD=45°$,$AD=BD=CD$,$AD⊥ BC$,$\therefore ∠ADF+∠FDC=90°$.
$\because DE⊥ DF$,$\therefore ∠EDA+∠ADF=90°$,$\therefore ∠EDA=∠FDC$.
在$△ AED$和$△ CFD$中,$\begin{cases} ∠EAD=∠FCD, \\ AD=CD, \\ ∠ADE=∠CDF, \end{cases}$
$\therefore △ AED≌△ CFD(\mathrm{ASA})$,
$\therefore AE=CF=5$,$DE=DF$,$\therefore △ EDF$为等腰直角三角形,
$\therefore ∠DEF=∠DFE=∠EDG=∠FDG=45°$,
$\therefore DG=EG=FG$,$\therefore DG=\dfrac{1}{2}EF$.
由(2)知$EF^2=BE^2+CF^2=144+25=169$,
$\therefore S_{△ DEF}=\dfrac{1}{2}EF· DG=\dfrac{1}{2}EF· \dfrac{1}{2}EF=\dfrac{1}{4}EF^2=\dfrac{169}{4}$.

第2题答图②