a,
$SD\subset (SCD)$
Trong $(ABCD)$, kẻ $AN\cap CD=K$
Mà $AN\subset (AMN), CD\subset (SCD), M\in (SCD)$
$\to (AMN)\cap (SCD)=KM$
Trong $(SCD)$, $SD\cap KM=I$
Mà $KM\subset (AMN)$
Vậy $SD\cap (AMN)=I$
b,
$N$ là trung điểm $OB$ nên $\dfrac{BN}{BO}=\dfrac{1}{2}$
$\to \dfrac{BN}{\dfrac{1}{2}BD}=\dfrac{1}{2}$
$\to \dfrac{BN}{BD}=\dfrac{1}{4}$
$\to \dfrac{BN+DN}{BN}=4$
$\to \dfrac{BN}{DN}=\dfrac{1}{3}$
Ta có $AB//DC$ nên theo Talet:
$\dfrac{KD}{AB}=\dfrac{DN}{AB}=3$
Mà $AB=CD\to \dfrac{KD}{CD}=3$
$\to \dfrac{CD}{KD}=\dfrac{1}{3}$
$\to \dfrac{KC}{KD}=\dfrac{2}{3}$
Áp dụng Menelaus vào $\Delta SCD$, cát tuyến $KMI$:
$\dfrac{KC}{KD}.\dfrac{MS}{MC}.\dfrac{ID}{IS}=1$
$\to \dfrac{2}{3}.1.\dfrac{ID}{IS}=1$
$\to \dfrac{ID}{IS}=\dfrac{3}{2}$
Vậy $\dfrac{SI}{ID}=\dfrac{2}{3}$
