$证明:(1)∵ 四边形ABCD是菱形,∴AB=AD,AC平分∠DAB,∴∠BAP =∠DAP$
$在△APB 和△APD 中$
$\begin{cases}{ AB=AD }\ \\ { ∠BAP=∠DAP } \\{ AP=AP} \end{cases}$
$∴△APB≌△APD(SAS)$
$(2)①∵四边形AP=AP, ABCD是菱形，∴AB//CD,AD//BC,AD=BC,∴△AFP∽△CBP$
$∴\frac{AF}{CB}=\frac{FP}{BP}$
$∵DF:FA=1:2,∴AF:BC=AF: AD=AF:(DF+AF)=2:3,∴\frac{FP}{BP}=\frac{2}{3}$
$由(1)，易知BP=PD=x,∵FP=y,∴\frac {y}{x}=\frac{2}{3},即y=\frac{2}{3}x$
$②当x=6时，y=\frac{2}{3}×6=4,∴FB=FP+PB=10$
$∵ DG//AB,∴△DFG∽△AFB,∴\frac{FG}{FB}=\frac{FD}{FA}=\frac{1}{2}$
$∴FG=\frac{1}{2}×10=5,∴线段FG的长为5$
