Giải thích các bước giải:
Ta có: $\Delta AMC,\Delta BMD$ đều
$\to CA=CM=MA,\widehat{AMC}=\widehat{CAM}=\widehat{ACM}=60^o$
$BM=MD=DB,\widehat{DMB}=\widehat{MBD}=\widehat{BDM}=60^o$
Xét $\Delta MCB,\Delta MAD$ có:
$MC=MA$
$\widehat{CMB}=180^o-\widehat{CMA}=180^o-60^o=180^o-\widehat{DMB}=\widehat{AMD}$
$MD=MB$
$\to\Delta CMB=\Delta AMD(c.g.c)$
$\to AD=BC,\widehat{ADM}=\widehat{CBM}$
Mà $E,F$ là trung điểm $AD,CB$
$\to DE=\dfrac12AD=\dfrac12BC=BF,\widehat{EDM}=\widehat{FBM}$
Xét $\Delta EDM,\Delta FBM$ có:
$DE=BF$
$\widehat{EDM}=\widehat{FBM}$
$MD=MB$
$\to\Delta EDM=\Delta FBM(c.g.c)$
$\to ME=MF,\widehat{EMD}=\widehat{FMB}$
$\to\widehat{EMF}=\widehat{EMD}+\widehat{DMF}=\widehat{FMB}+\widehat{DMF}=\widehat{DMB}=60^o$
$\to\Delta MEF$ đều
