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

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

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