解:仍然全等,理由如下:
∵$AD$和$A′D′$分别是$△ABC$和$△A′B′C′$的$BC$和$B′C′$边上的中线,
∴$BD = CD,$$B′D′ = C′D′.$
在$△ADC$和$△EDB$中,
$\{ \begin {array}{l}{AD = DE} \\{∠ADC = ∠BDE,} \\{BD = CD} \end {array} .$
∴$△ADC≌△EDB(\mathrm {SAS}).$
∴$AC = EB,$$∠DAC = ∠E,$
同理$A′C′ = E′B′,$$∠D′A′C′ = ∠E′.$
∵$AC = A′C′,$
∴$EB = E′B′.$
∵$AD = A′D′,$$AD = DE,$$A′D′ = D′E′,$
∴$AE = A′E′.$
∵$AB = A′B′,$
∴$△ABE≌△A′B′E′(\mathrm {SSS}).$
∴$∠BAE = ∠B′A′E′,$$∠E = ∠E′.$
∴$∠DAC = ∠D′A′C′.$
∴$∠BAC = ∠B′A′C′,$
又$AB = A′B′,$$AC = A′C′,$
∴$△ABC≌△A′B′C′(\mathrm {SAS});$