Giải thích các bước giải:
a.Ta có : $SH\perp BM, SA\perp (ABCD)\to SA\perp BH\to BH\perp (SAH)\to BH\perp AH$
$\to AH\perp BM$
b.Ta có : $AC\perp BC, \widehat{BAC}=30^o\to BC=\dfrac 12 AB=a, AC=a\sqrt{3}$
Mà $SA\perp ABCD\to SA\perp AB\to SB=AB\sqrt{2}=2a\sqrt{2}$
Do $\widehat{CAB}=30^o$
$\to\cos 30^o=\dfrac{MA^2+AB^2-MB^2}{2MA.AB}$
$\to \dfrac{\sqrt{3}}{2}=\dfrac{x^2+4a^2-MB^2}{2.x.2a}$
$\to x^2+4a^2-MB^2=2\sqrt{3}ax$
$\to MB^2=x^2+4a^2+2\sqrt{3}ax$
$\to MC^2=MB^2-BC^2=x^2+4a^2+2\sqrt{3}ax-a^2=x^2+3a^2+2\sqrt{3}ax$
$\to MC=x+a\sqrt{3}$
Lại có : $MH.MB=MA.MC$
$\to MH=\dfrac{x.(x+a\sqrt{3})}{\sqrt{x^2+4a^2+2\sqrt{3}ax}}$
$\to HB=MH+MB=\dfrac{x.(x+a\sqrt{3})}{\sqrt{x^2+4a^2+2\sqrt{3}ax}}+\sqrt{x^2+4a^2+2\sqrt{3}ax}$
$\to SH^2=SB^2-BH^2=2a\sqrt{2}-(\dfrac{x.(x+a\sqrt{3})}{\sqrt{x^2+4a^2+2\sqrt{3}ax}}+\sqrt{x^2+4a^2+2\sqrt{3}ax})^2$