Gọi $M$ là trung điểm $BC$
$\to \dfrac{d(G;(SCD))}{d(M;(SCD))}=\dfrac{2}{3}$
$\to d(G;(SCD))=\dfrac{2}{3}d(M;(SCD))$
Mà $d(M;(SCD))=\dfrac{1}{2}d(B;(SCD))$
$\to d(G;(SCD))=\dfrac{1}{3}d(B;(SCD))$
Mặt khác $AB//CD$ nên $d(B;(SCD))=d(A;(SCD))$
Kẻ $AH\bot SD$
Ta có $\begin{cases} CD\bot AD\\ CD\bot SA\end{cases}$
$\to CD\bot (SAD)$
$\to CD\bot AH$
Suy ra $AH\bot (SCD)$
$\to d(A;(SCD))=AH$
$\dfrac{1}{SA^2}+\dfrac{1}{AD^2}=\dfrac{1}{AH^2}$
$\to AH=\dfrac{a\sqrt{30}}{5}$
Vậy $d(G;(SCD))=\dfrac{AH}{3}=\dfrac{a\sqrt{30}}{15}$
