Cho tứ diện ABCD có \(AB \bot \left( {BCD} \right)\). Trong \(\Delta BCD\) vẽ các đường cao BE và DF cắt nhau ở O.

* Hướng dẫn giải

Ta có:

\(\left. \begin{array}{l} CD \bot BE\\ CD \bot AB \end{array} \right\} \Rightarrow \begin{array}{*{20}{c}} {}\\ {\left. \begin{array}{l} CD \bot \left( {ABE} \right)\\ CD \subset \left( {ADC} \right) \end{array} \right\} \Rightarrow \left( {ADC} \right) \bot \left( {ABE} \right)} \end{array}\)

Vậy “\(\left( {ADC} \right) \bot \left( {ABE} \right)\)”: ĐÚNG.

\(\left. \begin{array}{l} DF \bot BC\\ DF \bot AB \end{array} \right\} \Rightarrow {\left. \begin{array}{l} DF \bot \left( {ABC} \right)\\ SC \subset \left( {ABC} \right) \end{array} \right\}} \Rightarrow {\left. \begin{array}{l} DF \bot AC\\ DK \bot AC \end{array} \right\}} \Rightarrow {\left. \begin{array}{l} AC \bot \left( {DFK} \right)\\ AC \subset \left( {ADC} \right) \end{array} \right\}} \Rightarrow {\left( {ADC} \right) \bot \left( {DFK} \right)}\)

Vậy “ \(\left( {ADC} \right) \bot \left( {DFK} \right)\) ”: ĐÚNG.

Ta có

\(\left. \begin{array}{l} CD \bot BE\\ CD \bot AB \end{array} \right\} \Rightarrow {\left. \begin{array}{l} CD \bot \left( {ABE} \right)\\ CD \subset \left( {BDC} \right) \end{array} \right\} \Rightarrow } {\left( {BDC} \right) \bot \left( {ABE} \right)} \)

Vậy “ \(\left( {BDC} \right) \bot \left( {ABE} \right)\)”: ĐÚNG.

“ \(\left( {ADC} \right) \bot \left( {ABC} \right)\)”: SAI