在 △ABC\bigtriangleup {ABC}△ABC 中, DDD 为 BC{BC}BC 的中点,且 ∠DAC+∠BAC=π\angle {DAC} + \angle {BAC} = \pi∠DAC+∠BAC=π .(1)求 ABAD\frac{AB}{AD}ADAB ;(2)若 BC=22AC{BC} = {2\sqrt{2}}{AC}BC=22AC ,求 cosC\cos CcosC .