Giải thích các bước giải:
Ta có : $SA\perp (ABC)\to SA\perp BC$
Mà $BC\perp AC\to BC\perp (SAC)$
$\to V_{SABC}=\dfrac13BC.S_{SAC}=\dfrac{2a^3}{3}$
$\to BC.S_{SAC}=2a^3$
$\to BC.\dfrac12SA.AC=2a^3$
$\to a.\dfrac12SA.AC=2a^3\to SA.AC=4a^2$
Kẻ $AD\perp SC\to d(A,SBC)=AD$
Vì $SA\perp AC, AD\perp SC$
$\to \dfrac{1}{AD^2}=\dfrac{1}{SA^2}+\dfrac{1}{AC^2}$
$\to \dfrac{1}{AD^2}=\dfrac{SA^2+AC^2}{SA^2.AC^2}=\dfrac{SC^2}{(SA.AC)^2}=\dfrac{1}{4a^2}$
$\to AD=2a\to C$