a) Ta có:
$(SAD) \cap (SBC) = \left\{S\right\}$
Trong $mp(ABCD)$ gọi $AD\cap BC = \left\{E\right\}$
$E \in AD;\, AD \subset (SAD) \Rightarrow E\in (SAD)$
$E\in BC;\, BC \subset (SBC)\Rightarrow E \in (SBC)$
$\Rightarrow (SAD) \cap (SBC) = \left\{E\right\}$
$\Rightarrow (SAD) \cap (SBC) = SE$
b) Ta có:
$(MAB) \cap (SAC) = \left\{A\right\}\quad (1)$
Bên cạnh đó:
$(SAC) \cap (SBD) = \left\{S\right\}$
Trong $mp(ABCD)$ gọi $AC\cap BD = \left\{F\right\}$
$F\in AC;\, AC\subset (SAC)\Rightarrow F\in (SAC)$
$F\in BD;\, BD\subset (SBD)\Rightarrow F\in (SBD)$
$\Rightarrow (SAC) \cap (SBD) = \left\{F\right\}$
$\Rightarrow (SAC) \cap (SBD) =SF$
Trong $mp(SBD)$ gọi $SF\cap BM = \left\{I\right\}$
$I\in SF;\, SF\in (SAC)\Rightarrow I\in (SAC)$
$I\in BM;\, BM\subset (MAB)\Rightarrow I \in (MAB)$
$\Rightarrow (MAB) \cap (SAC) = \left\{I\right\}\quad (2)$
$(1)(2)\Rightarrow (MAB) \cap (SAC) = AI$
