证明：$(1)$在$\triangle ABD$和$\triangle CDB$中
$\begin {cases}AD = CB\\AB = CD\\BD = DB\end {cases}$
∴$\triangle ABD≌\triangle CDB(\mathrm {SSS})$
∴$∠ADB=∠CBD，$∴$AD// BC$
解：$(2)$由$(1)$得$AD//BC，$∴$∠EDG=∠F BG$
情况$1$：当点$E$沿$D→A(0\leq t\leq 3)$运动时，
$DE = 4t，$$CF = t$
又$AD = BC = 12，$∴$BF = 12 - t$
$①$若$\triangle DEG≌\triangle BFG$
则$DE = BF，$$DG = BG$
∴$4t = 12 - t，$解得$t = 2.4$
又$BD = 15，$∴此时$BG=\frac 12BD = 7.5$
$②$若$\triangle DEG≌\triangle BGF$
则$DE = BG，$$DG = BF$
∴$DE + BF = BG + DG = BD$
即$4t + 12 - t = 15，$解得$t = 1$
此时$BG = DE = 4×1 = 4；$
情况$2$：当点$E$沿$A→ D(3运动时，$
$DE = 12×2 - 4t = 24 - 4t，$$CF = t$
则$BF = 12 - t$
$①$若$\triangle DEG≌\triangle BFG$
则$DE = BF，$$DG = BG$
∴$24 - 4t = 12 - t，$解得$t = 4$
此时$BG=\frac 12BD = 7.5；$
$②$若$\triangle DEG≌\triangle BGF$
则$DE = BG，$$DG = BF$
∴$DE + BF = BD$
即$24 - 4t+12 - t = 15，$解得$t = 4.2$
此时$BG = DE = 24 - 4×4.2 = 7.2$
综上，$\triangle DEG $和$\triangle BFG $全等时的情况如下
当$t = 2.4，$$BG = 7.5$或$t = 4，$$BG = 7.5$时，
$\triangle DEG≌\triangle BFG；$
当$t = 1，$$BG = 4$或$t = 4.2，$$BG = 7.2$时，
$\triangle DEG≌\triangle BGF$
