Ta có:
$G\in (GEF)$
$G\in (BCD)$
$\Rightarrow (GEF)\cap (BCD)=\left\{G\right\}$
Trong mp$(ABD)$, gọi $EF\cap BD = \left\{K\right\}$
$K\in EF;\, EF\subset (GEF)\Rightarrow K\in (GEF)$
$K\in BD;\, BD\subset (BCD)\Rightarrow K\in (BCD)$
$\Rightarrow (GEF)\cap (BCD)=\left\{K\right\}$
$\Rightarrow (GEF)\cap (BCD)=GK$