Ta có:
$SH\perp (ABC)\quad (gt)$
$AC\subset (ABC)$
$\to SH\perp AC$
Lại có: $HK\perp AC\quad (gt)$
$\to AC\perp (SHK)$
mà $HT\subset (SHK)$
nên $AC\perp HT$
Ta được:
$\left.\begin{array}{l}HT \perp SK\quad (gt)\\HT\perp AC\quad (cmt)\\SK\subset (SAC)\\AC\subset (SAC)\end{array}\right\}\longrightarrow HT\perp (SAC)\quad (đpcm)$