a,
$AD\subset (ACD)$
Trong $(BCD)$ kẻ $MN\cap CD=E$
$\to (MNP)\cap (ACD)=PE$
Trong $(ACD)$ kẻ $PE\cap AD=Q$
$\to AD\cap (MNP)=Q$
b,
Áp dụng Menelaus vào $\Delta BCD$, cát tuyến $ENM$:
$\dfrac{ED}{EC}.\dfrac{NB}{ND}.\dfrac{MC}{MB}=1$
$\to \dfrac{ED}{EC}.1.3=1$
$\to \dfrac{ED}{EC}=\dfrac{1}{3}$
Áp dụng Menelaus vào $\Delta ACD$, cát tuyến $EQP$:
$\dfrac{ED}{EC}.\dfrac{AQ}{QD}.\dfrac{PC}{PA}=1$
$\to \dfrac{1}{3}.\dfrac{AQ}{QD}.2=1$
$\to \dfrac{AQ}{QD}=\dfrac{3}{2}$