$证明：(1) ∵四边形ABCD是平行四边形$ $∴AD=CB,∠A=∠C,∠ADC=∠ABC$ $∵ DE平分∠ADC,BF平分∠ABC$ $∴∠ADE=\frac 12∠ADC,∠CBF= \frac 12∠ABC$ $∴∠ADE =∠CBF$ $在△ADE和△CBF中$ $\begin{cases}∠A=∠C \\AD=CB\\∠ ADE=∠CBF\end{cases}$ $∴△ADE≌△CBF$ $∴DE=BF$
$证明:(2)如图,连接EF$ $∵四边形ABCD是平行四边形$ $∴AB //CD, AB = CD$ $∴∠BEF=∠DFE$ $∵ DE//BF$ $∴∠BFE=∠DEF$ $在△BEF和△DFE中$ $\begin{cases}∠ BEF=∠DFE\\EF=FE\\∠BFE=∠DEF\end{cases}$ $△BEF≌△DFE$ $∴ BE=DF$ $∴AB- BE=CD-DF ,即AE=CF$
$解:(3)成立$ $∵四边形ABCD是平行四边形$ $∴AD=CB，AB=CD,∠A=∠C$ $∵BE=DF$ $∴AB- BE=CD- DF，即AE= CF$ $在△ADE和△CBF中$ $\begin{cases}AD=CB\\∠A=∠C\\AE=CF\end{cases}$ $∴△ADE≌△CBF$ $∴DE= BF$
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