$?證明:∵AB//DE,BC//EF,∴∠A=∠EDF,∠BCA=∠F?$ $?∵AD=CF,∴AC=DF?$ $?在△ABC和△DEF中?$ $?{{\begin{cases} {{∠A=∠EDF}} \\ {AC=DF} \\ {∠BCA=∠F} \end{cases}}}?$ $?∴△ABC≌△DEF(ASA)?$ $?證明:延長AE到點F使得EF=AE,連接DF?$ $?在△ABE和△FDE中?$ $?{{\begin{cases} {{BE=DE}} \\ {∠AEB=∠FED} \\ {AE=FE} \end{cases}}}?$ $?∴△ABE≌△FDE(SAS)?$ $?∴AB=FD,∴∠ABE=∠FDE?$ $?∴FD=DC,∠FDE+∠ADE=∠ABE+∠BAD=∠ADC?$ $?在△ADF和△ADC中?$ $?{{\begin{cases} {{AD=AD}} \\ {∠ADF=∠ADC} \\ {DF=DC} \end{cases}}}?$ $?∴△ADF≌△ADC(SAS)?$ $?∴AF=AC?$ $?∴AC=AE+EF=2AE?$
$解:當∠MCN+α=180°時,BE=CF,證明:$ $∵∠CBE+∠BCE=180°-α$ $∠FCA+∠BCE=∠BCA=∠MCN=180°-α$ $∴∠CBE=∠FCA$ $在△BCE和△CAF中$ ${{\begin{cases} {{∠CBE=∠ACF}} \\ {∠BEC=∠CFA} \\ {BC=CA} \end{cases}}}$ $∴△BCE≌△CAF(AAS)$ $∴BE=CF$
$解:EF=BE+AF,證明:$ $由三角形內角和知∠ACF+∠CAF=180°-α$ $由E,C,F三點共線知∠ACF+∠BCE=180°-α$ $∴∠CAF=∠BCE$ $在△BEC和△CFA中$ ${{\begin{cases} {{∠BEC=∠CFA}} \\ {∠BCE=∠CAF} \\ {BC=CA} \end{cases}}}$ $∴△BEC≌△CFA(AAS)$ $∴EC=FA,BE=CF$ $∴EF=EC+CF=AF+BE$
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