解答:
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(Ⅰ)证明:∵BC∥平面ADE,BC?平面PBC,
平面PBC∩平面ADE=DE,
∴BC∥DE.
∵D为PC中点,
∴E为PB的中点.
(Ⅱ)解:∵AP=AB,E为PB的中点,∴AE⊥PB,
又PB⊥AD,∴PB⊥平面ADE,
得DE⊥PB,且平面PBC⊥平面ADE.
由BC∥DE,得BC⊥PB.
过C作CH⊥ED于H,由平面PBC⊥平面ADE,∴CH⊥平面ADE.
∴∠CAH是直线AC与平面ADE所成的角.
∵BC∥DE,BC⊥PB,∴
CH=BE=PB=,
∴
sin∠CAH==.