Giải thích các bước giải:
a.Gọi E là trung điểm MD $\to N= (E, EM)\cap AC$
Vì C,D đối xứng qua AB $\to\Delta ABD$ đều
Gọi $DG\perp BC=G\to BG=BM=\dfrac 12, \widehat{DBG}=60^o\to GD=\dfrac{\sqrt{3}}{2}$
$\to DM=\dfrac{\sqrt{7}}{2}$
Mà $\widehat{DMN}=180^o-\widehat{DAN}=60^o\to MN=\dfrac 12 DM=\dfrac{\sqrt{7}}{4}$
$\to DN=\dfrac{\sqrt{21}}{4}$
$\to S_{DMN}=\dfrac{1}{2}.DN.MN=\dfrac{7\sqrt 3}{32}$
b.Ta có : $\widehat{MNP}=\widehat{ADM}=\arctan\dfrac{AM}{AD}=\arctan \dfrac{\sqrt{3}}{2}$
$\to \dfrac{MP}{NP}= \dfrac{\sqrt{3}}{2}$
$\to NP=MP.\dfrac{2}{\sqrt 3}$
$\to S_{MNP}=\dfrac 12NP. MP=\dfrac{MP^2}{\sqrt 3}=\dfrac{\sqrt 3}{16}$ vì $MP\perp AC$
c.Ta có: $\widehat{DBG}=60^o=\widehat{ACB}\to DB//AC\to MP\perp BD\to \cos(MD, PD)=0$