Cho hình chóp S.ABCD có đáy ABCD là hình chữ nhật...

Gọi H là trung điểm của AB, do tam giác SAB đều $\Rightarrow SH\bot AB$.
Ta có: $\left\{ \begin{aligned}
& \left( SAB \right)\cap \left( ABC\text{D} \right)=AB \\
& \left( SAB \right)\bot \left( ABC\text{D} \right) \\
& \left( SAB \right)\supset SH\bot AB \\
\end{aligned} \right.\Rightarrow SH\bot \left( ABC\text{D} \right)$

Ta có: $AH\cap \left( S\text{D}B \right)\Rightarrow B\Rightarrow \dfrac{d\left( A;(S\text{D}B) \right)}{d\left( H;(S\text{D}B) \right)}=\dfrac{AB}{HB}=2\Rightarrow d\left( A;(S\text{D}B) \right)=2\text{d}\left( H;(S\text{D}B) \right)$
Trong $\left( ABC\text{D} \right)$ kẻ $HM\bot B\text{D }\left( M\in B\text{D} \right)$, trong $\left( SHM \right)$ kẻ $HK\bot SM\left( K\in SM \right)$
Ta có: $\left\{ \begin{aligned}
& B\text{D}\bot HM \\
& B\text{D}\bot \text{SH }\left( SH\bot (ABC\text{D}) \right) \\
\end{aligned} \right.\Rightarrow B\text{D}\bot \left( SHM \right)\Rightarrow B\text{D}\bot HK$
$\left\{ \begin{aligned}
& HK\bot SM \\
& HK\bot B\text{D} \\
\end{aligned} \right.\Rightarrow HK\bot \left( S\text{D}B \right)\Rightarrow d\left( H;(S\text{D}B) \right)=HK$
Trong $\left( ABC\text{D} \right)$ kẻ $A\text{E}\bot B\text{D }\left( E\in B\text{D} \right)\Rightarrow A\text{E // HM}$
Ta có $A\text{E}=\dfrac{AB.A\text{D}}{\sqrt{A{{B}^{2}}+A{{\text{D}}^{2}}}}=\dfrac{2\text{a}.a}{\sqrt{4{{\text{a}}^{2}}+{{a}^{2}}}}=\dfrac{2\text{a}}{\sqrt{5}}$
Có HM là đường trung bình của tam giác ABE $\Rightarrow HM=\dfrac{1}{2}A\text{E}=\dfrac{a}{\sqrt{5}}$
Tam giác SAB đều cạnh $AB=2\text{a}\Rightarrow SH=\dfrac{2\text{a}\sqrt{3}}{2}=a\sqrt{3}$
Xét tam giác vuông SHM: $HK=\dfrac{SH.HM}{\sqrt{S{{H}^{2}}+H{{M}^{2}}}}=\dfrac{a\sqrt{3}.\dfrac{a}{\sqrt{5}}}{\sqrt{3{{\text{a}}^{2}}+\dfrac{{{a}^{2}}}{5}}}=\dfrac{a\sqrt{3}}{4}$
Vậy $d\left( A;(S\text{D}B) \right)=2.\dfrac{a\sqrt{3}}{4}=\dfrac{a\sqrt{3}}{2}$.