Question

数学求救

Answer 1

(1)∵ PA⊥底面ABCD，故AP⊥改侍CD.
又 CD⊥AC，从而CD⊥面ACP.
∵AE在面ACP上，故知CD⊥AE.

(2) PA⊥底核御吵面ABCD => AP⊥AB
又AB⊥AD，从而AB⊥面PAD => PD⊥AB.
AB=BC，∠ABC=60° => △ABC为等边三角形.

=> AC=AB=PA，即有△APC为等腰三角形.
又E为PC的中点，故AE⊥PC.
由（1）可知，拆颂AE垂直CD => AE⊥面PCD =>AE⊥PD
从而知，PD⊥面ABE.

Answer 2

(1) PA⊥面ABCD =>PA⊥CD
CD⊥AC =>CD⊥面PAC =>CD垂直AE
(2) AB=BC
∠ABC=60 =>AB=AC=BC=PA
又E是PC终点 =>AE⊥核数陆PC
加上（1）中AE⊥改顷CD
所毕弊以AE⊥面PCD =>PD⊥AE
又PA⊥AB AB⊥AD =>AB⊥面PAD =>AB⊥PD
综上有PD⊥面ABE

Answer 3

(1)PA⊥面ABCD =>PA⊥CD
AC⊥CD =>CD⊥面PAC =>CD垂直AE
(2)AE⊥CD
AB=BC，∠ABC=60 =>ABC为等边三槐灶键角形=>AC=AB

PA=AB=AC,PA⊥AC=>ACP为等辩弯腰直角铅巧三角形
E为PC中点=>AE⊥PC
=>AE⊥面PCD =>AE⊥PD
AB⊥PD=>PD⊥面ABE

Answer 4

数学题别问我。