(1)∵AB=
1
3 AC,
∴BC=2AB,
∵线段AB长为x,
∴BC=2x,
∵A为CD的中点,
∴AD=AC=3AB,
∴AD=3x,
故答案为:2x,3x;
(2)∵点P为BD的中点,
∴BP=PD=4cm,
∵AB=x,AD=3x,
∴AD+AB=DB=8,
即x+3x=8,
∴x=2,
∴DC=6x=12.
(1)∵AB=
1
3 AC,
∴BC=2AB,
∵线段AB长为x,
∴BC=2x,
∵A为CD的中点,
∴AD=AC=3AB,
∴AD=3x,
故答案为:2x,3x;
(2)∵点P为BD的中点,
∴BP=PD=4cm,
∵AB=x,AD=3x,
∴AD+AB=DB=8,
即x+3x=8,
∴x=2,
∴DC=6x=12.