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 $$ PDC $$

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 $$  PDC $$

Uniteasy Admin · Accepted Answer

(1) In $$ △PCD $$ , make parallel line to $$ CD $$ through point $$ E $$ .
$$ \because AB⊥ $$ plane $$ PBC $$ , $$ AB \parallel CD $$
$$ 	herefore EF ⊥ $$ plane $$ PBC $$ , $$ EF⊥BF $$
$$ \because E $$ is the midpoint of $$ PD $$
$$ 	herefore EF = \dfrac{1}{2} DC = AB $$
$$ 	herefore ABFE $$ 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 $$ PDC $$

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