$(2)解:由(1)得,△ABC≌△ADC$ $∴AB=AD=4,BC=DC=3$ $∴S_{四边形ABCD}=S_{△ABC}+S_{△ADC}=\frac {1}{2}×3×4×2=12$
(2)解:不一定成立,理由如下: 如图,可与(1)同理证得△BAE≌△ACF ∴BE=AF,AE=CF ∴AF=AE+EF 即BE=CF+EF
$证明:∵AB//DE$ $∴∠A=∠EDF$ $在△ABC和△DEF中$ ${{\begin{cases} {{∠A=∠EDF}} \\ {∠B=∠E} \\ {BC=EF} \end{cases}}}$ $∴△ABC≌△DEF(AAS)$ $∴AC=DF$ $∴AC-DC=DF-DC$ $即AD=CF$
$证明:∵AC平分∠BAD$ $∴∠BAC=∠DAC$ $在△ABC和△ADC中$ ${{\begin{cases} {{∠B=∠D}} \\ {∠BAC=∠DAC} \\ {AC=AC} \end{cases}}}$ $∴△ABC≌△ADC(AAS)$
$解:①能判定,证明:$ $如图$ $在△ABD和△A'B'D'中$ ${{\begin{cases} {{∠B=∠B'}} \\ {∠ADB=∠A'D'B'} \\ {AD=A'D'} \end{cases}}}$ $∴△ABD≌△A'B'D'(AAS)$ $∴AB=A'B'$ $在△ABC和△A'B'C'中$ ${{\begin{cases} {{∠C=∠C'}} \\ {∠B=∠B'} \\ {AB=A'B'} \end{cases}}}$ $∴△ABC≌△A'B'C'(AAS)$
$②不能判定,如图$
$③不能判定,如图$
$解:BE+EF=CF,理由如下:$ $∵∠BAE+∠FAC=90°,$ $∠ACF+∠FAC=90°$ $∴∠BAE=∠ACF$ $在△BAE和△ACF中$ ${{\begin{cases} {{∠BEA=∠AFC}} \\ {∠BAE=∠ACF} \\ {AB=CA} \end{cases}}}$ $∴△BAE≌△ACF(AAS)$ $∴BE=AF,AE=CF$ $∴EF+FA=AE$ $即EF+BE=CF$
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