a,
$SD\subset (SCD)$
Trong $(ABCD)$, kẻ $AN\cap CD=K$
Mà $\begin{cases} AN\subset (AMN)\\ CD\subset (SCD)\\ M\in(SCD)\end{cases}$
$\to (AMN)\cap (SCD)=KM$
Trong $(SCD)$, kẻ $SD\cap MK=I$
Mà $MK\subset (AMN)$
Vậy $SD\cap (AMN)=I$
b,
$N$ là trung điểm $BO$ nên:
$\dfrac{NB}{ND}=\dfrac{\dfrac{1}{2}OB}{BD-BN}=\dfrac{ \dfrac{1}{2}BO}{ 2BO-\dfrac{1}{2}BO}=\dfrac{1}{3}$
Mà $AB//CD$ nên theo Talet, ta có:
$\dfrac{KD}{AB}=\dfrac{ND}{NB}=3$
Mặt khác $AB=CD$ nên $\dfrac{KD}{CD}=3$
$\to \dfrac{CD}{KD}=\dfrac{1}{3}$
$\to \dfrac{KC}{KD}=\dfrac{2}{3}$
Áp dụng Menelaus cho $\Delta SCD$, cát tuyến $KMI$:
$\dfrac{KC}{KD}.\dfrac{MS}{MC}.\dfrac{ID}{IS}=1$
$\to \dfrac{2}{3}.1.\dfrac{ID}{IS}=1$
$\to \dfrac{ID}{IS}=\dfrac{3}{2}$
$\to \dfrac{SI}{ID}=\dfrac{2}{3}$
Vậy $\dfrac{SI}{SD}=\dfrac{2}{5}$
