Question Number 213330 by MrGaster last updated on 02/Nov/24
Answered by MrGaster last updated on 03/Nov/24
![My answer to the first question: (1)PA⊥ABCD→PA⊥AD+PB⊥AD→AD⊥PAB→AD⊥AB+AB⊥BC+ABCD→AD//BC→AD//PBC [Q.E.D] (2)……](https://www.tinkutara.com/question/Q213360.png)
$$\mathrm{My}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{first}\:\mathrm{question}: \\ $$$$\left(\mathrm{1}\right){PA}\bot{ABCD}\rightarrow{PA}\bot{AD}+{PB}\bot{AD}\rightarrow{AD}\bot{PAB}\rightarrow{AD}\bot{AB}+{AB}\bot{BC}+{ABCD}\rightarrow{AD}//{BC}\rightarrow{AD}//{PBC} \\ $$$$\left[{Q}.{E}.{D}\right] \\ $$$$\left(\mathrm{2}\right)\ldots\ldots \\ $$