$AB=CD=a\\ AD=BC=a\sqrt{2}\\ AC=\sqrt{AB^2+BC^2}=a\sqrt{3}\\ AM=\dfrac{\sqrt{2}}{2}\\ BM=\sqrt{AB^2+AM^2}=a\dfrac{\sqrt{6}}{2}\\ \cos(\widehat{CAB})=\dfrac{CA^2+AB^2-BC^2}{2CA.AB}=\dfrac{\sqrt{3}}{3}\\ \cos(\widehat{ABM})=\dfrac{AB^2+BM^2-AM^2}{2AB.BM}=\dfrac{\sqrt{6}}{3}\\ \widehat{AIB}=180^o-\widehat{CAB}-\widehat{ABM}\\ =180^o-\arccos\left(\dfrac{\sqrt{3}}{3}\right)-\arccos\left(\dfrac{\sqrt{6}}{3}\right)\\ =90^o\\ \Rightarrow MB \perp AC\\ \text{Mà} \,\,MB \perp SA(SA \perp (ABCD))\\ \Rightarrow MB \perp (SAC)\\ \Rightarrow (SMB) \perp (SAC)$