Kẻ $AH\bot BD, AK\bot SH$
Ta có: $BD\bot AH, BD\bot SA$ vì $SA\bot(ABCD)$
$\to BD\bot(SAH)$
$\to BD\bot AK$
Suy ra $AK\bot(SBD)$
$\to d(A,(SBD))=AK$
$\Delta ABD$: $\dfrac{1}{AB^2}+\dfrac{1}{AD^2}=\dfrac{1}{AH^2}$
$\to \Delta SAH$: $\dfrac{1}{AK^2}=\dfrac{1}{SA^2}+\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}+\dfrac{1}{AD^2}$
$\to AK=\dfrac{3a\sqrt{13}}{13}$
Vậy $d(A,(SBD))=\dfrac{3a\sqrt{13}}{13}$
