In Rt△ABC, ∠ c = 90, cosB=BCAB=35, let BC=3x, then AB=5x.

AC=AB2？ BC2=4x，

At Rt△HBC, cosB=BHBC=35, BC=3x,

∴BH=95x，

∵Rt△ABC rotates around vertex C to get RT △ A ′ B ′ C, where point B ′ falls right on AB.

∴ca′=ca=4x,cb′=cb,∠a′=∠a，

∵ch⊥bb′，

∴b′h=bh=95x，

∴ab′=ab-b′h-bh=75x，

∠∠ADB′=∠A′DC，∠A′=∠A，

∴△adb′∽△a′dc，

∴AB'A'C=B'DDC, that is, 75x4x=B'DDC,

∴b′ddc=720.

So the answer is 720.