在四棱锥P-ABCD中CD//AB,AD⊥AB,AD=DC=1/2AB,BC⊥PC,（1）求证PA⊥BC

（1）∵AC=√（AD^2+DC^2）=√2/2AB,作CE⊥AB,同理可得BC=√2/2AB,AC^2+BC^2=1/2AB^2+1/2AB^2=AB^2∴BC⊥AC∵BC⊥PC,PC∩AC于C∴BC⊥平面PAC∴PA⊥BC（2）作PA中点N,连结MNDC∵M,N分别为PA,PB中点∴MN‖=1/2AB∵CD‖=1/2AB∴四边形MNDC为平行四边形∴CM平行DN∵DN∈平面PAD,CM∉平面PAD∴CM‖平面PAD