Question

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

Answer 1

(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=

.