如图,在四棱锥 P−ABCDP - {ABCD}P−ABCD 中,底面四边形 ABCD{ABCD}ABCD 满足 AB=CB=2,AD=CD=5,∠ABC=90∘{AB} = {CB} = \sqrt{2},{AD} = {CD} = \sqrt{5},\angle {ABC} = {90}^{ \circ }AB=CB=2,AD=CD=5,∠ABC=90∘ ,棱 PD{PD}PD 上的点 EEE 满足 PE=2DE{PE} = {2DE}PE=2DE . (1) 证明:直线 CE//{CE}//CE// 平面 PAB{PAB}PAB ;(2) 若 PB=5,PD=22{PB} = \sqrt{5},{PD} = {2\sqrt{2}}PB=5,PD=22 ,且 PA=PC{PA} = {PC}PA=PC ,求直线 CE{CE}CE 与平面 PBC{PBC}PBC 所成角的正弦值.