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?$證明?：(1)?∵將?ABCD?沿過(guò)點(diǎn)?A?的直線?l?折疊,$?
?$使點(diǎn)?D?落到?AB?邊上的點(diǎn)?D′?處,$?
?$∴?∠DAE=∠D′AE，∠DEA=∠D′EA，∠D=∠AD′E，?$?
?$∵?DE∥AD′，?$?
?$∴?∠DEA=∠EAD′，?$?
?$∴?∠DAE=∠EAD′=∠DEA=∠D′EA，?$?
?$∴?∠DAD′=∠DED′，?$?
?$∴四邊形?DAD′E?是平行四邊形,$?
?$∴?DE=AD′，?$?
?$∵四邊形?ABCD?是平行四邊形,$?
?$∴?AB∥DC?且?AB=DC，?$?
?$∴?CE∥D′B?且?CE=D′B，?$?
?$∴四邊形?BCED′?是平行四邊形.$?
?$?(2)?∵?BE?平分?∠ABC，?$?
?$∴?∠CBE=∠EBA，?$?
?$∵?AD∥BC，?$?
?$∴?∠DAB+∠CBA=180°，?$?
?$∵?∠DAE=∠BAE，?$?
?$∴?∠EAB+∠EBA=90°，?$?
?$∴?∠AEB=90°.?$?

?$ 證明:(1)∵四邊形ABCD是平行四邊形$?
?$∴AD//BC,AB//CD,AB=CD$?
?$∴∠AEB=∠DAE$?
?$∵AE是∠BAD的平分線$?
?$∴∠BAE=∠DAE$?
?$∴∠BAE=∠AEB$?
?$∴AB=BE$?
?$∴∠AEB=∠EAB=65°$
$∴∠BAD=∠BAE+∠EAD=130°?$
?$(2)∵AB=BE,BF⊥AE$?
?$∴AF=EF$?
?$∵AD//BC$?
?$∴∠ADF=∠ECF,∠DAF=∠AEC$?
?$在△ADF和△ECF中$?
?$\begin{cases}{∠ADF=∠ECF}\\{∠DAF=∠AEC}\\{AF=EF}\end{cases}$?
?$∴△ADF≌△ECF$?
?$∴CF=DF$?
?$∵AF=EF,CF=DF$?
?$∴四邊形ACED是平行四邊形。$?
?$解:如圖，?BE=CG，??GH=5m，??EF=11m；?$?
?$根據(jù)題意可知：?△CHG∽△CAB，?$?
?$?△CFE∽△CAB，?$?
?$則有：?\frac {CG}{CB}=\frac {HG}{AB}，\frac {CE}{CB}=\frac {EF}{AB}?$?
?$設(shè)?BE=CG=x，BC=y?$?
?$則?\frac {x}{y}=\frac {5}{AB}，\frac {y-x}{y}=\frac {11}{AB}?$?
?$所以兩式相加,得?\frac {y}{y}=\frac {16}{AB}?$?
?$所以?AB=16m?$?
?$所以她說(shuō)得對(duì)$?
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