Đáp án:

$d\left( {BC,SA} \right) = \dfrac{{4a}}{5}$

Giải thích các bước giải:

Kẻ tia $Ax//BC$ và $HD\bot Ax=D$. Kẻ $HK\bot SD=K$

Ta có:

$\begin{array}{l}
d\left( {BC,SA} \right)\\
 = d\left( {BC,\left( {SAD} \right)} \right)\left( {do:BC//AD} \right)\\
 = d\left( {B,\left( {SAD} \right)} \right)\\
 = 2d\left( {H,\left( {SAD} \right)} \right)\left( {do:BA \cap \left( {SAD} \right) = A;BA = 2HA} \right)
\end{array}$

Mà ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
AD \bot HD\\
AD \bot SH\\
HD \cap SH = H
\end{array} \right.\\
 \Rightarrow AD \bot \left( {SHD} \right)\\
 \Rightarrow AD \bot HK
\end{array}$

Nên:

$\begin{array}{l}
\left\{ \begin{array}{l}
HK \bot AD\\
HK \bot SD\\
AD \cap SD = D
\end{array} \right.\\
 \Rightarrow HK \bot \left( {SAD} \right)\\
 \Rightarrow d\left( {H,\left( {SAD} \right)} \right) = HK\\
 \Rightarrow d\left( {BC,SA} \right) = 2HK
\end{array}$

Ta có:

$\begin{array}{l}
 + )\Delta AHC;\widehat A = {90^0};AH = \dfrac{1}{2}AB = a;AC = a\\
 \Rightarrow HC = AC\sqrt 2  = a\sqrt 2 \\
 + )\Delta SHC;\widehat H = {90^0};HC = a\sqrt 2 ;SC = \dfrac{{a\sqrt {70} }}{5}\\
 \Rightarrow SH = \sqrt {S{C^2} - H{C^2}}  = \dfrac{{2a}}{{\sqrt 5 }}
\end{array}$

Lại có:

$\begin{array}{l}
 + )\Delta ABC;\widehat A = {90^0};AB = 2a;AC = a\\
 \Rightarrow BC = \sqrt {A{B^2} + A{C^2}}  = a\sqrt 5 \\
 \Rightarrow \sin \widehat {ABC} = \dfrac{{AC}}{{BC}} = \dfrac{1}{{\sqrt 5 }}\\
 + )\Delta AHD;\widehat D = {90^0};AH = a\\
 \Rightarrow HD = AH\sin \widehat {DAH}\\
 \Rightarrow HD = AH\sin \widehat {ABC}\left( {do:\widehat {DAH} = \widehat {ABC}\left( {slt} \right)} \right)\\
 \Rightarrow HD = \dfrac{a}{{\sqrt 5 }}
\end{array}$

Như vậy:

$\begin{array}{l}
\Delta SHD;\widehat H = {90^0};SH = \dfrac{{2a}}{{\sqrt 5 }};HD = \dfrac{a}{{\sqrt 5 }};HK \bot SD = K\\
 \Rightarrow HK = \sqrt {\dfrac{1}{{\dfrac{1}{{S{H^2}}} + \dfrac{1}{{H{D^2}}}}}}  = \dfrac{{2a}}{5}\\
 \Rightarrow d\left( {BC,SA} \right) = 2HK = \dfrac{{4a}}{5}\\
 \Rightarrow d\left( {BC,SA} \right) = \dfrac{{4a}}{5}
\end{array}$

Vậy $d\left( {BC,SA} \right) = \dfrac{{4a}}{5}$