a,
$ABCD$ hình vuông nên $BD\bot AC$
$SA\bot(ABCD)\to BD\bot SA$
Suy ra $BD\bot(SAC)$
Vậy $(SBD)\bot(SAC)$
b,
$SA\bot(ABCD)\to (SH,(ABCD))=(SH,AH)$
$AB=a\to AC=a\sqrt2$
$\to SA=\sqrt{SC^2-AC^2}=a\sqrt3$
$AB=a, BH=\dfrac{BC}{2}=\dfrac{a}{2}\to AH=\sqrt{AB^2+BH^2}=\dfrac{a\sqrt5}{2}$
$\to \tan\widehat{SHA}=\dfrac{SA}{AH}=\dfrac{2\sqrt{15}}{3}$
Vậy $\tan(SH,(ABCD))=\dfrac{2\sqrt{15}}{3}$
c,
Gọi $P$ là trung điểm $CD$
$\to BD//(HKP)$
$\to d(HK,BD)=d(BD,(HKP))=d(O,(HKP))$
Gọi $I=HP\cap OC$
Có $OC\bot BD, HP//BD$ nên $OC\bot HP=I$
$HI//BO\to \dfrac{CI}{CO}=\dfrac{CH}{CB}=\dfrac{1}{2}$
$\to \dfrac{IO}{IA}=\dfrac{1}{3}$
$\to d(O,(HKP))=\dfrac{d(A,(HKP))}{3}$
Kẻ $AH\bot KI$
$HP\bot AI, HP\bot AK$ nên $HP\bot(KIA)$
$\to HP\bot AH$
$\to AH\bot(KHP)$
$\to d(A,(KHP))=AH$
$AK=\dfrac{SA}{2}=\dfrac{a\sqrt3}{2}$
$AI=\dfrac{3}{4}AC=\dfrac{3a\sqrt2}{4}$
$\dfrac{1}{AH^2}=\dfrac{1}{AK^2}+\dfrac{1}{AI^2}$
$\to AH=\dfrac{3a\sqrt5}{10}$
Vậy $d(HK,BD)=\dfrac{a\sqrt5}{10}$
