Cho tứ diện ABCD có các tam giác ABC và DBC...

- Hướng dẫn giải

Đáp án là B.

$BC=AB\sqrt{2}=2a\sqrt{2}$.Gọi  H là trung điểm BC  ta có:

$\left\{\begin{array}{l}AH\perp BC\\ BC=\left(ABC\right)\cap \left(DBC\right)\\ \left(ABC\right)\perp \left(DBC\right)\end{array}\right.\Rightarrow AH\perp \left(DBC\right)$

kẻ $HE\perp DC$, $HK\perp AE$ (1)

$\begin{array}{l}\left\{\begin{array}{l}DC\perp HE\\ DC\perp AH(doAH\perp \left(DBC\right)\subset DC)\end{array}\right.\\ \Rightarrow DC\perp \left(AHE\right)\Rightarrow DC\perp HK\left(2\right)\end{array}$

từ $\left(1\right)\&\left(2\right)$ $HK\perp \left(ADC\right)\Rightarrow d\left(H;\left(ADC\right)\right)=HK$

$$\begin{gathered} d\left(B;\left(ADC\right)\right)=2d\left(H;\left(ADC\right)\right) \\ =\frac{2AH.HE}{\sqrt{A{H}^2+H{E}^2}}=\frac{2\sqrt{6}}{3} \end{gathered}$$

$AH=\frac{BC}{2},HE=\frac{AB}{2}$;$AH=\frac{BC}{2}=a\sqrt{2},HE=\frac{BC}{2}=a$