如图,在三棱锥P-ABC中,PA⊥底面ABC,∠BCA=90°,AP=AC,点D,E分别在棱PB,PC上,且BC∥平面ADE.(Ⅰ

2025-05-11 15:26:10
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回答1:

(Ⅰ)∵BC∥平面ADE,BC?平面PBC,平面PBC∩平面ADE=DE
∴BC∥ED.
∵PA⊥底面ABC,BC?底面ABC,
∴PA⊥BC.
又∠BCA=90°,∴AC⊥BC.
∵PA∩AC=A,∴BC⊥平面PAC.
∴DE⊥平面PAC.
(Ⅱ)由(Ⅰ)知,DE⊥平面PAC,
∵PC?平面PAC,∴DE⊥PC,
又∵PC⊥AD,AD∩DE=D,∴PC⊥平面ADE,∴AE⊥PC,
∵AP=AC,∴E是PC的中点,ED是△PBC的中位线.
 

VP?ABC
VP?ADE
=
S△PBC
S△PED
=
4
1

∴VABCED=
3
4
VP?ABC
=
3
4
×8
=6.