如图所示,矩形ABCD中,AB=a,AD=b,过点D作DE⊥AC于E,交直线AB于F.现将△ACD沿对角线AC折起到△P

(1)见解析 (2)


(1)证明:∵DF⊥AC,
∴折起后AC⊥PE,AC⊥EF,
∴AC⊥平面PEF,
又PH⊂平面PEF,
∴AC⊥PH,
又PH⊥EF,EF∩AC=E,
∴PH⊥平面ABC.
(2)解:∵PE⊥AC,EF⊥AC,
∴∠PEF就是二面角P AC B的平面角,
∴∠PEF=60°,
∴Rt△PHE中,PH= PE,
折起前,Rt△ADC中,
DE= = ,
S _△ABC = ab,
折起后,PE=DE,
∴PH= PE= · ,
∴ = PH·S _△ABC
= · · · ab
= · ,
∵a+b=2,a>0,b>0,
∴ ≤ = ≤ = ,
当且仅当a=b=1时,两个等号同时成立,
因此( ) _max = .