Đáp án: $C$
Giải thích các bước giải:
Ta có $M, N, P, Q$ là trung điểm $AB, BC, CD, DA$
$\to MN, NP, PQ, QM$ là đường trung bình $\Delta ABC, \Delta BCD, \Delta CDA,\Delta DBA$
$\to MN//AC//PQ, QM//BD//PN$
Xét $\Delta BMN, \Delta BAC$ có:
Chung $\hat B$
$\widehat{BMN}=\widehat{BAC}$ vì $MN//AC$
$\to \Delta BMN\sim\Delta BAC(g.g)$
$\to \dfrac{S_{BMN}}{S_{ABC}}=(\dfrac{BM}{BA})^2=\dfrac14$
$\to S_{BMN}=\dfrac14S_{BAC}$
Tương tự chứng minh được $S_{CNP}=\dfrac14S_{CBD}, S_{DPQ}=\dfrac14S_{ADC},S_{AQM}=\dfrac14S_{ABD}$
$\to S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ}=\dfrac14S_{ABD}+\dfrac14S_{BAC}+\dfrac14S_{CBD}+\dfrac14S_{ACD}$
$\to S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ}=\dfrac14(S_{ABD}+S_{BAC}+S_{CBD}+S_{ACD})$
$\to S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ}=\dfrac14((S_{ABD}+S_{CBD})+(S_{BAC}+S_{ACD}))$
$\to S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ}=\dfrac14(S_{ABCD}+S_{ABCD})$
$\to S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ}=\dfrac12S_{ABCD}$
$\to S_{ABCD}-( S_{AQM}+S_{BMN}+S_{CNP}+S_{DPQ})=\dfrac12S_{ABCD}$
$\to S_{MNPQ}=\dfrac12S_{ABCD}$