Question

In the pyramid P-ABCD , △PBC is a right triangle, AB⊥ plane PBC , AB \parallel CD , AB = \dfrac{1}{2} DC . E is the midpoint of PD . Prove:

(1) AE \parallel plane PBC .

(2) AE ⊥ plane PDC

Steven Zheng Steven Zheng posted 2 months ago


Answer

(1) In △PCD , make parallel line to CD through point E .

\because AB⊥ plane PBC , AB \parallel CD

\therefore EF ⊥ plane PBC , EF⊥BF

\because E is the midpoint of PD

\therefore EF = \dfrac{1}{2} DC = AB

\therefore ABFE is a rectangle

AE \parallel plane PBC

(2) \because E is the midpoint of PD, EF \parallel CD

F is also the midpoint of PC

\because △PBC is right triangle,

BF ⊥ PC

\because BF⊥ plane PDC

AE⊥ plane PDC

Steven Zheng Steven Zheng posted 2 months ago

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