Hình bạn tự vẽ
a) Xét $\triangle$ ACE và $\triangle$ ABD có
AC = AB (gt)
AE = AD ( gt)
$\widehat{CAE}$ = $\widehat{BAD}$ $(=60^{o})$
$\Rightarrow$ $\triangle$ ACE = $\triangle$ ABD (c.g.c)
$\Rightarrow$ BD = CE
b)Từ $\triangle$ ACE = $\triangle$ ABD $\Rightarrow$ $\widehat{ACE}$ = $\widehat{ABD}$
Ta có $\widehat{BMC}$ = $180^o$ - ($\widehat{MBC}$ + $\widehat{MCB}$ )
= $180^o$ - ($60^o$ - $\widehat{ABD}$ + $60^o$ + $\widehat{ACE}$ ) mà $\widehat{ABD}$ = $\widehat{ACE}$
$\Rightarrow$ $\widehat{BMC}$ = $180^o$ - ($60^o$ + $60^o$) = $180^o$ - $120^o$ = $60^o$
c)MB = BD + DM = CE +MD = CM + ME +MD = (ME+MD) + CM = MA + MC
d) suy ra từ câu c) ta có MA = ME + MD