$AD\subset (ACD)$
Trong $(BCD)$, $NP\cap CD=I$
Mà $NP\subset (MNP), CD\subset (ACD)$
$\Rightarrow (MNP)\cap (ACD)=MI$
$MI\cap AD=Q$
$\Rightarrow AD\cap (MNP)=Q$
Ta có $\dfrac{NB}{NC}=1; \dfrac{PD}{PB}=\dfrac{1}{2}$
Áp dụng Menelaus vào $\Delta BCD$, cát tuyến $IPB$:
$\dfrac{IC}{ID}.\dfrac{PD}{PB}.\dfrac{NB}{NC}=1$
$\Leftrightarrow \dfrac{IC}{ID}=2$
$\Leftrightarrow \dfrac{ID}{IC}=\dfrac{1}{2}$
Áp dụng Menelaus vào $\Delta ACD$, cát tuyến $IQM$:
$\dfrac{ID}{IC}.\dfrac{QA}{QD}.\dfrac{MC}{MA}=1$
$\Leftrightarrow \dfrac{QA}{QD}=2$
$\Leftrightarrow \dfrac{AQ}{AD}=\dfrac{2}{3}$