Giải thích các bước giải:

a.Xét $\Delta MAD<\Delta MCB$ có:

$MA=MC$
$\widehat{AMD}=180^o-\widehat{DMB}=180^o-\widehat{CMA}=\widehat{BMC}$

$MD=MB$

$\to \Delta AMD=\Delta CMB(c.g.c)$

$\to AD=CB$

b.Từ a $\to \widehat{MAD}=\widehat{MCB}$

$\to \widehat{MAE}=\widehat{MCF}$

Mà $AE=\dfrac12AD=\dfrac12BC=CF$

$\to \Delta MAE=\Delta MCF(c.g.c)$

$\to ME=MF,\widehat{AME}=\widehat{CMF}$

$\to \widehat{EMF}=\widehat{CMF}+\widehat{CME}=\widehat{AME}+\widehat{CME}=\widehat{CMA}=60^o$

$\to \Delta MEF$ đều